本文探讨了三个具体的算法挑战，包括从混淆的电话号码中提取数字、在损坏的比分板上推测最接近的比赛结果以及识别学术会议中最可能被伪造的主题。通过深入分析和提供源代码，展示了如何运用算法来解决这些问题。
Getting the Digits
Problem
You just made a new friend at an international puzzle conference, and you asked for a way to keep in touch. You found the following note slipped under your hotel room door the next day:
“Salutations, new friend! I have replaced every digit of my phone number with its spelled-out uppercase English representation (“ZERO”, “ONE”, “TWO”, “THREE”, “FOUR”, “FIVE”, “SIX”, “SEVEN”, “EIGHT”, “NINE” for the digits 0 through 9, in that order), and then reordered all of those letters in some way to produce a string S. It’s up to you to use S to figure out how many digits are in my phone number and what those digits are, but I will tell you that my phone number consists of those digits in nondecreasing order. Give me a call… if you can!”
You would to like to call your friend to tell him that this is an obnoxious way to give someone a phone number, but you need the phone number to do that! What is it?
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int n;
char S[10][10]={"ZERO", "ONE", "TWO", "THREE", "FOUR", "FIVE", "SIX", "SEVEN", "EIGHT", "NINE"};
char s[10000];
int a[26]={0};
void work(int i,int j) {
int m=strlen(S[i]);
Rep(k,m) {
a[S[i][k]-'A']-=j;
}
}
int t[10]={0};
int main()
{
freopen("A.in","r",stdin);
freopen("a.out","w",stdout);
int T;
cin>>T;
For(kcase,T)
{
scanf("%s",s);
n=strlen(s);
sort(s,s+n);
MEM(a)
Rep(i,n) a[s[i]-'A']++;
MEM(t)
int p1,p2;
p1=0;p2='Z';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=2;p2='W';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=4;p2='U';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=8;p2='G';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=6;p2='X';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=5;p2='F';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=1;p2='O';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=3;p2='H';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=7;p2='V';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
p1=9;p2='I';
t[p1]=a[p2-'A'];
work(p1,t[p1]);
printf("Case #%d: ",kcase);
Rep(i,10) Rep(j,t[i]) cout<<i;
cout<<endl;
}
return 0;
}
Close Match
You are attending the most important game in sports history. The Oceania Coders are playing the Eurasia Jammers in the Centrifugal Bumble-Puppy world finals. Unfortunately, you were sleep deprived from all the anticipation, so you fell asleep during the game!
The scoreboard is currently displaying both scores, perhaps with one or more leading zeroes (because the scoreboard displays a fixed number of digits). While you were asleep, some of the lights on the scoreboard were damaged by strong ball hits, so one or more of the digits in one or both scores are not being displayed.
You think close games are more exciting, and you would like to imagine that the scores are as close as possible. Can you fill in all of the missing digits in a way that minimizes the absolute difference between the scores? If there is more than one way to attain the minimum absolute difference, choose the way that minimizes the Coders’ score. If there is more than one way to attain the minimum absolute difference while also minimizing the Coders’ score, choose the way that minimizes the Jammers’ score.
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
int n;
string s1,s2,ans1,ans2;
#define LL long long
#define ULL unsigned long long
#define LD long double
#define Rep(x,y,i) for (int i=x;i<y;i++) //[x,y)
#define RepD(x,y,i) for (int i=x;i>y;i--) //(y,x]
#define Mem(X) memset(X,0,sizeof(X));
#define Pr(X) cout<<" "<<#X<<"="<<X<<" ";
#define PrL(X) cout<<#X<<" = "<<X<<endl;
#define PrLL cout<<endl;
#define IntMod 10000
const double EPS = 1e-10;
struct BigInt
{
private:
vector<int> A;
bool Positive;
int VecNum;
public:
//Constructors
inline int GetLength(LL a) {int t=0; while (a>0) {a/=IntMod; t++;} return t;}
BigInt() { Positive = 1; VecNum = 0; }
BigInt(const BigInt &a) { A = a.A; VecNum = a.VecNum; Positive = a.Positive; }
BigInt(string s)
{
A.reserve(100);
if (s == "-0") { A.push_back(0); Positive = 1; VecNum = 1; return; }
int kk = 3; int kl = 0;
int l = s.length();
int j = l - 1;
if (s[0] == '-') { Positive = 0; kk++; kl++; }
else Positive = 1;
while (j>kk)
{
int t = 0;
Rep(0, 4, i) { t *= 10; t += s[j - (3 - i)] - '0'; }
j -= 4;
A.push_back(t);
}
int t = 0;
int k = j + 1;
Rep(kl, k, i) { t *= 10; t += s[i] - '0'; }
A.push_back(t);
VecNum = A.size();
}
BigInt(const LL &b)
{
LL a=b;
Positive=(a>=0);
VecNum=GetLength(abs(a));
A.resize(VecNum);
Rep(0,VecNum,i)
{
A[i]=a%IntMod;
a/=IntMod;
}
}
//BigInt& operator = (BigInt& a) { VecNum=a.VecNum; Positive=a.Positive; A=a.A; }
BigInt& operator = (const string s) { BigInt x(s); *this=x; }
BigInt& operator = (LL s) { BigInt x(s); *this = x; }
//Basic Math Functions
friend BigInt abs(BigInt a) { a.Positive=1; return a; }
friend bool isnegative(BigInt &a) { return a.Positive; }
BigInt& operator - () { Positive = !(Positive); return *this; }
// Ostream and Instream
friend ostream& operator << (ostream &out, BigInt &a)
{
if (a.VecNum==0)
{
out<<"0";
return out;
} //Bug Fixed if there is a BigInt constructed by the default constructor
if (!a.Positive) out << "-";
out << a.A[a.VecNum - 1];
RepD(a.VecNum - 2, -1, i)
{
if (a.A[i] == 0) { out << "0000"; continue; }
Rep(0, (int)(4 - log(a.A[i]) / log(10) - EPS), j) out << '0';
out << a.A[i];
}
return out;
}
friend istream& operator >> (istream &in, BigInt &a)
{
string s;
in >> s;
int L = s.length() - 1;int i = 0;bool flag = 0;
if (s[i] == '-') { i++; flag = 1; }
while (s[i] == '0' && i < L) i++;
string b(s.begin() + i, s.end());
if (flag) b.insert(0, "-");
a = BigInt(b);
return in;
}
//Bool Operators
bool operator < (const BigInt &b) const
{
if (Positive && b.Positive)
{
if (VecNum != b.VecNum) return (bool)(VecNum<b.VecNum);
RepD(VecNum-1, -1, i)
if (A[i] != b.A[i])
return (bool)(A[i]<b.A[i]);
//Bug Fixed that there should be a RepD rather that Rep
return 0;
}
if (!Positive && b.Positive) return 1;
if (Positive && !b.Positive) return 0;
BigInt a = b;
BigInt c = (*this);
if (!c.Positive && !a.Positive) return !((-c) < (-a));
return 1;
}
bool operator > (const BigInt &b) const { return !((*this)<b); }
bool operator == (const BigInt &b) const { return (Positive == b.Positive && A == b.A && VecNum == b.VecNum); }
bool operator <= (const BigInt &b) const { return (*this == b) || (*this < b); }
bool operator >= (const BigInt &b) const { return (*this == b) || !(*this < b);}
bool operator != (const BigInt &b) const { return !(*this == b); }
bool operator < (const string b) const { BigInt x(b); return (*this) < x; }
bool operator == (const string b) const { BigInt x(b); return x==(*this) < x; }
bool operator < (const LL& b) const { BigInt x(b); return *this < x; }
bool operator == (const LL& b) const { BigInt x(b); return *this == x; }
//Function of Plus and Minus
BigInt operator - (const BigInt &b) const
{
BigInt x=*this;
BigInt y=b;
if (!x.Positive && y.Positive) return -(-x + y);
if ( x.Positive && !y.Positive) return x + y;
if (!x.Positive && !y.Positive) return (-y) - (-x);
if (x<y) return -(y - x);
int L = max(x.VecNum, y.VecNum);
y.A.resize(L);
x.A.resize(L);
Rep(0, L, i)
{
x.A[i] -= y.A[i];
if (x.A[i]<0) { x.A[i] += IntMod; x.A[i + 1]--; }
}
while (x.A[L - 1] == 0 && ((L-1)!=0) ) { x.VecNum--; x.A.pop_back(); L--; }
return x;
}
BigInt operator + (const BigInt &b) const
{
BigInt x=*this;
BigInt y=b;
if (!x.Positive && y.Positive) return y - (-x);
if (!x.Positive && !y.Positive) return -(-x + (-y));
if ( x.Positive && !y.Positive) return x - (-y);
int L = max(x.VecNum, y.VecNum);
x.A.resize(L + 1);
y.A.resize(L + 1);
for (int i = 0; i<L; i++) x.A[i] += y.A[i];
for (int i = 0; i<L; i++)
{ x.A[i + 1] += x.A[i] / IntMod; x.A[i] %= IntMod; }
x.VecNum = L;
if (x.A[L]) x.VecNum++;
else x.A.erase(x.A.begin() + L);
return x;
}
BigInt operator - (const LL &b) const { BigInt y(b); return *this - y; }
BigInt operator + (const LL &b) const { BigInt y(b); return *this + y; }
BigInt operator - (const string b) const { BigInt y(b); return *this - y; }
BigInt operator + (const string b) const { BigInt y(b); return *this + y; }
BigInt operator += (const BigInt& b) { *this=*this+b; return *this; }
BigInt operator += (const LL& b) { *this=*this+b; return *this; }
BigInt operator -= (const BigInt& b) { *this=*this-b; return *this; }
BigInt operator -= (const LL& b) { *this=*this-b; return *this; }
//Function of Multiply and Division
BigInt operator * (const BigInt& b) const
{
BigInt x=*this;
BigInt y=b;
BigInt ans;
if ((x.Positive && y.Positive) || (!x.Positive && !y.Positive)) ans.Positive=1;
else ans.Positive=0;
int m=x.VecNum,n=y.VecNum;
int L=m+n+1;
ans.VecNum=L;
ans.A.resize(L+3);
Rep(0,m,i)
Rep(0,n,j)
{
int pos=i+j;
LL t=x.A[i]*y.A[j];
int post=0;
while (t>0)
{
ans.A[pos+post]+=t%IntMod;
t/=IntMod;
post++;
}
}
Rep(0,L,i) { ans.A[i + 1] += ans.A[i] / IntMod; ans.A[i] %= IntMod; }
while (ans.A[L - 1] == 0 && ((L-1)!=0) ) { ans.VecNum--; ans.A.pop_back(); L--; }
return ans;
}
BigInt operator * (const LL &b) const { BigInt x(b); return (*this) * x; }
BigInt operator * (const string b) const { BigInt x(b); return (*this) * x; }
BigInt operator *= (const BigInt& b){ *this = *this * b; return *this; }
BigInt operator *= (const LL &b) { *this = *this * b; return *this; }
BigInt operator *= (const string b) { *this = *this * b; return *this; }
BigInt operator / (const BigInt& b) const
{
}
//Function of Remainder
BigInt operator % (const BigInt& b) const
{
}
//Function of BITS
};
bool fff=0;
void check(string p1,string p2) {
// cout<<p1<<' '<<p2<<endl;
if (!fff) {
fff=1;
ans1=p1,ans2=p2;
return;
}
BigInt ppp=abs(BigInt(p1)-BigInt(p2));
BigInt pp2=abs(BigInt(ans1)-BigInt(ans2));
if (ppp<pp2) {
fff=1;
ans1=p1,ans2=p2;
return;
} else if (ppp==pp2&&BigInt(p1)<BigInt(ans1)){
fff=1;
ans1=p1,ans2=p2;
return;
} else if (ppp==pp2&&BigInt(p1)==BigInt(ans1)&&BigInt(p2)<BigInt(ans2) ) {
fff=1;
ans1=p1,ans2=p2;
return;
}
}
void fi(string &s,int i,int j,int p) {
Fork(k,i,j) if (s[k]=='?')s[k]=p;
}
int main()
{
freopen("B-large.in","r",stdin);
freopen("Bbb.out","w",stdout);
int T;
cin>>T;
For(kcase,T)
{
fff=0;
bool f=0;
cin>>s1>>s2;
n=s1.length();
string p1=s1,p2=s2;
for(int i=0;i<n;i++) {
char p='0';
if (s1[i]=='?'&&s2[i]=='?') {
s1[i]=s2[i]='0';
p1=s1,p2=s2;
p1[i]='1';
fi(p1,i+1,n-1,'0');
fi(p2,i+1,n-1,'9');
check(p1,p2);
p1=s1,p2=s2;
p1[i]='0'; p2[i]='1';
fi(p1,i+1,n-1,'9');
fi(p2,i+1,n-1,'0');
check(p1,p2);
}
else if (s1[i]=='?'||s2[i]=='?')
{
if (s1[i]=='?') {
p1=s1,p2=s2;
if (s2[i]!='0') {
p1[i]=p2[i]-1;
fi(p1,i+1,n-1,'9');
fi(p2,i+1,n-1,'0');
check(p1,p2);
}
p1=s1,p2=s2;
if (s2[i]!='9') {
p1[i]=p2[i]+1;
fi(p1,i+1,n-1,'0');
fi(p2,i+1,n-1,'9');
check(p1,p2);
}
}
else {
p1=s1,p2=s2;
if (s1[i]!='0') {
p2[i]=p1[i]-1;
fi(p2,i+1,n-1,'9');
fi(p1,i+1,n-1,'0');
check(p1,p2);
}
p1=s1,p2=s2;
if (s1[i]!='9') {
p2[i]=p1[i]+1;
fi(p2,i+1,n-1,'0');
fi(p1,i+1,n-1,'9');
check(p1,p2);
}
}
p=s1[i]=='?' ?s2[i]:s1[i];
s1[i]=s2[i]=p;
}
else if (s1[i]==s2[i]) continue;
else if (s1[i]!=s2[i]) {
f=1;
if (s1[i]<s2[i]) {
fi(s1,i+1,n-1,'9');
fi(s2,i+1,n-1,'0');
check(s1,s2);
}
else {
fi(s1,i+1,n-1,'0');
fi(s2,i+1,n-1,'9');
check(s1,s2);
}break;
}
}
if (!f) check(s1,s2);
printf("Case #%d: ",kcase);
cout<<ans1<<' '<<ans2;
cout<<endl;
}
return 0;
}
Technobabble
Every year, your professor posts a blank sign-up sheet for a prestigious scientific research conference on her door. If a student wants to give a lecture at the conference, they choose a two-word topic that is not already on the sheet and write it on the sheet. Once the deadline has passed, the professor has one of her grad students put the topics in a random order, to avoid biasing for or against students who signed up earlier. Then she presents the topics to you for review.
Since the snacks at the conference are excellent, some students try to fake their way into the conference. They choose the first word of some topic already on the sheet and the second word of some topic already on the sheet, and combine them (putting the first word first, and the second word second) to create a new “topic” (as long as it isn’t already on the sheet). Since your professor is open-minded, sometimes this strategy actually works!
The fakers are completely unoriginal and can’t come up with any new first or second words on their own; they must use existing ones from the sheet. Moreover, they won’t try to use an existing first word as their own second word (unless the word also already exists on the sheet as a second word), or vice versa.
You have a list of all N of the submitted topics, in some arbitrary order; you don’t know the order in which they were actually written on the sheet. What’s the largest number of them that could have been faked?
题意：给一个二分图，取尽量少的边，覆盖所有点
先求二分图匹配，然后剩下的点每个各取一条边
#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (2010)
pair<string ,string> p[MAXN];
#define MAXM (7000)
class Max_flow //dinic+当前弧优化
{
public:
int n,s,t;
int q[MAXN];
int edge[MAXM],next[MAXM],pre[MAXN],weight[MAXM],size;
void addedge(int u,int v,int w)
{
edge[++size]=v;
weight[size]=w;
next[size]=pre[u];
pre[u]=size;
}
void addedge2(int u,int v,int w){addedge(u,v,w),addedge(v,u,0);}
bool b[MAXN];
int d[MAXN];
bool SPFA(int s,int t)
{
For(i,n) d[i]=INF;
MEM(b)
d[q[1]=s]=0;b[s]=1;
int head=1,tail=1;
while (head<=tail)
{
int now=q[head++];
Forp(now)
{
int &v=edge[p];
if (weight[p]&&!b[v])
{
d[v]=d[now]+1;
b[v]=1,q[++tail]=v;
}
}
}
return b[t];
}
int iter[MAXN];
int dfs(int x,int f)
{
if (x==t) return f;
Forpiter(x)
{
int v=edge[p];
if (weight[p]&&d[x]<d[v])
{
int nowflow=dfs(v,min(weight[p],f));
if (nowflow)
{
weight[p]-=nowflow;
weight[p^1]+=nowflow;
return nowflow;
}
}
}
return 0;
}
int max_flow(int s,int t)
{
int flow=0;
while(SPFA(s,t))
{
For(i,n) iter[i]=pre[i];
int f;
while (f=dfs(s,INF))
flow+=f;
}
return flow;
}
void mem(int n,int s,int t)
{
(*this).n=n;
(*this).t=t;
(*this).s=s;
size=1;
MEM(pre)
}
}S;
vector<string> h1,h2;
int find(string s,vector<string>& h) {
return lower_bound(h.begin(),h.end(),s)-h.begin();
}
int main()
{
freopen("C.in","r",stdin);
freopen("c.out","w",stdout);
int T;
cin>>T;
For(kcase,T)
{
h1.clear();h2.clear();
int n=read();
Rep(i,n) {
cin>>p[i].fi>>p[i].se;
h1.pb(p[i].fi);
h2.pb(p[i].se);
}
sort(h1.begin(),h1.end());
sort(h2.begin(),h2.end());
int m1=unique(h1.begin(),h1.end())-h1.begin();
int m2=unique(h2.begin(),h2.end())-h2.begin();
For(i,n-m1) h1.pop_back();
For(i,n-m2) h2.pop_back();
int s=m1+m2+1,t=m1+m2+2;
// cout<<h1.size()<<' '<<h2.size()<<endl;
S.mem(t,s,t);
For(i,m1) S.addedge2(s,i,1);
For(i,m2) S.addedge2(i+m1,t,1);
Rep(i,n) {
S.addedge2(find(p[i].fi,h1)+1,find(p[i].se,h2)+1+m1,1);
// cout<<find(p[i].fi,h1)+1<<' '<<find(p[i].se,h2)+1+m1<<endl;
}
int c=S.max_flow(s,t);
int lef=m1+m2-c*2;
int ans=n-(lef+c);
// cout<<n<<' '<<c<<' '<<m1+m2<<' '<<m1<<' '<<m2<<endl;
printf("Case #%d: %d\n",kcase,ans);
}
return 0;
}
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