本文提供ACM-ICPC2018焦作赛区网络预赛的题解，涵盖从A到K等多道题目，包括MagicMirror、MathematicalCurse、JiuYuanWantstoEat等，解析了算法思路并提供了代码实现。
这场打得还是比较爽的，但是队友差一点就再过一题，还是难受啊。
每天都有新的难过
A. Magic Mirror
Jessie has a magic mirror.
Every morning she will ask the mirror: 'Mirror mirror tell me, who is the most beautiful girl in the world?' If the mirror says her name, she will praise the mirror: 'Good guy!', but if the mirror says the name of another person, she will assail the mirror: 'Dare you say that again?'
Today Jessie asks the mirror the same question above, and you are given a series of mirror's answers. For each answer, please output Jessie's response. You can assume that the uppercase or lowercase letters appearing anywhere in the name will have no influence on the answer. For example, 'Jessie' and 'jessie' represent the same person.
Input
The first line contains an integer T(1 \le T \le 100)T(1≤T≤100), which is the number of test cases.
Each test case contains one line with a single-word name, which contains only English letters. The length of each name is no more than 1515.
Output
For each test case, output one line containing the answer.
样例输入复制
2 Jessie Justin
样例输出复制
Good guy! Dare you say that again?
题目来源
大小写转换一下
#include<stdio.h> #include<bits/stdc++.h> using namespace std; #define ll long long #define lson l,(l+r)/2,rt<<1 #define rson (l+r)/2+1,r,rt<<1|1 #define dbg(x) cout<<#x<<" = "<< (x)<< endl #define pb push_back #define fi first #define se second #define sz(x) (int)(x).size() #define pll pair<long long,long long> #define pii pair<int,int> #define pq priority_queue const int N=1e5+5,MD=1e9+7,INF=0x3f3f3f3f; const ll LL_INF=0x3f3f3f3f3f3f3f3f; const double eps=1e-9,e=exp(1),PI=acos(-1.); int a[N]; int main() { ios::sync_with_stdio(false),cin.tie(0),cout.tie(0); int T; cin>>T; while(T--) { string s; cin>>s; for(int i=0;s[i];i++)if(s[i]>='A'&&s[i]<='Z')s[i]+=32; if(s=="jessie")cout<<"Good guy!\n"; else cout<<"Dare you say that again?\n"; } return 0; }
B. Mathematical Curse
A prince of the Science Continent was imprisoned in a castle because of his contempt for mathematics when he was young, and was entangled in some mathematical curses. He studied hard until he reached adulthood and decided to use his knowledge to escape the castle.
There are NN rooms from the place where he was imprisoned to the exit of the castle. In the i^{th}ith room, there is a wizard who has a resentment value of a[i]a[i]. The prince has MM curses, the j^{th}jthcurse is f[j]f[j], and f[j]f[j] represents one of the four arithmetic operations, namely addition('+'), subtraction('-'), multiplication('*'), and integer division('/'). The prince's initial resentment value is KK. Entering a room and fighting with the wizard will eliminate a curse, but the prince's resentment value will become the result of the arithmetic operation f[j]f[j] with the wizard's resentment value. That is, if the prince eliminates the j^{th}jth curse in the i^{th}ithroom, then his resentment value will change from xx to (x\ f[j]\ a[i]x f[j] a[i]), for example, when x=1, a[i]=2, f[j]=x=1,a[i]=2,f[j]='+', then xxwill become 1+2=31+2=3.
Before the prince escapes from the castle, he must eliminate all the curses. He must go from a[1]a[1] to a[N]a[N] in order and cannot turn back. He must also eliminate the f[1]f[1] to f[M]f[M] curses in order(It is guaranteed that N\ge MN≥M). What is the maximum resentment value that the prince may have when he leaves the castle?
Input
The first line contains an integer T(1 \le T \le 1000)T(1≤T≤1000), which is the number of test cases.
For each test case, the first line contains three non-zero integers: N(1 \le N \le 1000), M(1 \le M \le 5)N(1≤N≤1000),M(1≤M≤5)and K(-1000 \le K \le 1000K(−1000≤K≤1000), the second line contains NN non-zero integers: a[1], a[2], ..., a[N](-1000 \le a[i] \le 1000)a[1],a[2],...,a[N](−1000≤a[i]≤1000), and the third line contains MM characters: f[1], f[2], ..., f[M](f[j] =f[1],f[2],...,f[M](f[j]='+','-','*','/', with no spaces in between.
Output
For each test case, output one line containing a single integer.
样例输入复制
3 2 1 5 2 3 / 3 2 1 1 2 3 ++ 4 4 5 1 2 3 4 +-*/
样例输出复制
2 6 3
题目来源
有一个负的，还有一个正的，所以都要进入dp
#include <bits/stdc++.h> using namespace std; typedef long long ll; const ll INF=1LL<<62; ll dp1[1005][6],dp2[1005][6]; int a[1005]; char s[10]; ll calc(ll x,char c,int y) { if(c=='-') return x-y; if(c=='+') return x+y; if(c=='/') return x/y; if(c=='*') return x*y; } int main() { int T,n,m,k; scanf("%d",&T); while(T--) { scanf("%d%d%d",&n,&m,&k); for(int i=0;i<=n;i++) for(int j=0;j<=m;j++) dp1[i][j]=-INF,dp2[i][j]=INF; for(int i=1;i<=n;i++) scanf("%d",&a[i]); for(int i=0;i<=n;i++) dp1[i][0]=dp2[i][0]=k; scanf("%s",s+1); for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) { dp1[i][j]=dp1[i-1][j]; dp2[i][j]=dp2[i-1][j]; if(dp1[i-1][j-1]!=-INF) dp1[i][j]=max(dp1[i][j],calc(dp1[i-1][j-1],s[j],a[i])); if(dp2[i-1][j-1]!=INF) dp1[i][j]=max(dp1[i][j],calc(dp2[i-1][j-1],s[j],a[i])); if(dp1[i-1][j-1]!=-INF) dp2[i][j]=min(dp2[i][j],calc(dp1[i-1][j-1],s[j],a[i])); if(dp2[i-1][j-1]!=INF) dp2[i][j]=min(dp2[i][j],calc(dp2[i-1][j-1],s[j],a[i])); } printf("%lld\n",dp1[n][m]); } return 0; }
E. Jiu Yuan Wants to Eat
You ye Jiu yuan is the daughter of the Great GOD Emancipator. And when she becomes an adult, she will be queen of Tusikur, so she wanted to travel the world while she was still young. In a country, she found a small pub called Whitehouse. Just as she was about to go in for a drink, the boss Carola appeared. And ask her to solve this problem or she will not be allowed to enter the pub. The problem description is as follows:
There is a tree with nn nodes, each node iicontains weight a[i]a[i], the initial value of a[i]a[i] is 00. The root number of the tree is 11. Now you need to do the following operations:
1)1) Multiply all weight on the path from uuto vv by xx
2)2) For all weight on the path from uu to vv, increasing xx to them
3)3) For all weight on the path from uu to vv, change them to the bitwise NOT of them
4)4) Ask the sum of the weight on the path from uu to vv
The answer modulo 2^{64}264.
Jiu Yuan is a clever girl, but she was not good at algorithm, so she hopes that you can help her solve this problem. Ding\backsim\backsim\backsim∽∽∽
The bitwise NOT is a unary operation that performs logical negation on each bit, forming the ones' complement of the given binary value. Bits that are 00become 11, and those that are 11 become 00. For example:
NOT 0111 (decimal 7) = 1000 (decimal 8)
NOT 10101011 = 01010100
Input
The input contains multiple groups of data.
For each group of data, the first line contains a number of nn, and the number of nodes.
The second line contains (n - 1)(n−1)integers b_ibi, which means that the father node of node (i +1)(i+1) is b_ibi.
The third line contains one integer mm, which means the number of operations，
The next mm lines contain the following four operations:
At first, we input one integer opt
1)1) If opt is 11, then input 33 integers, u, v, xu,v,x, which means multiply all weight on the path from uu to vv by xx
2)2) If opt is 22, then input 33 integers, u, v, xu,v,x, which means for all weight on the path from uu to vv, increasing xx to them
3)3) If opt is 33, then input 22 integers, u, vu,v, which means for all weight on the path from uu to vv, change them to the bitwise NOT of them
4)4) If opt is 44, then input 22 integers, u, vu,v, and ask the sum of the weights on the path from uu to vv
1 \le n,m,u,v \le 10^51≤n,m,u,v≤105
1 \le x < 2^{64}1≤x<264
Output
For each operation 44, output the answer.
样例输入复制
7 1 1 1 2 2 4 5 2 5 6 1 1 1 6 2 4 5 6 3 5 2 4 2 2 2 1 4 3 1 2 4 1 2 3 1 1 4 1 1
样例输出复制
5 18446744073709551613 18446744073709551614 0
题目来源
队友LCT差一点调出来唉
划重点 按位取反是*-1+MAX
#include<iostream> #include<cstdio> #include<cstring> #define N 100010 using namespace std; typedef unsigned long long ull; ull st[N],n,m,q,x,y,op,w,rev[N],size[N],c[N][2],fa[N]; ull mt[N],a[N],v[N],s[N]; bool isroot(ull x) { return c[fa[x]][0]!=x&&c[fa[x]][1]!=x; } void updata(ull x) { ull l=c[x][0],r=c[x][1]; size[x]=size[l]+size[r]+1; s[x]=s[l]+s[r]+v[x]; } void paint(ull x,ull cc,ull aa) { if (!x) return; v[x]=v[x]*cc+aa; s[x]=s[x]*cc+size[x]*aa; a[x]=a[x]*cc+aa; mt[x]=mt[x]*cc; } void pushdown(ull x) { ull l=c[x][0],r=c[x][1]; if (rev[x]) { rev[l]^=1; rev[r]^=1; rev[x]^=1; swap(c[x][0],c[x][1]); } ull aa=a[x],cc=mt[x]; a[x]=0; mt[x]=1; if (cc!=1||aa!=0) { paint(l,cc,aa); paint(r,cc,aa); } } void rotata(ull x) { ull y=fa[x],z=fa[y],l,r; if (x==c[y][0]) l=0; else l=1; r=l^1; if (!isroot(y)) { if (c[z][0]==y)c[z][0]=x; else c[z][1]=x; } fa[x]=z; fa[y]=x; fa[c[x][r]]=y; c[y][l]=c[x][r]; c[x][r]=y; updata(y); updata(x); } void splay(ull x) { ull top(0); st[++top]=x; for(ull i=x; !isroot(i); i=fa[i]) st[++top]=fa[i]; for (ull i=top; i; i--) pushdown(st[i]); while (!isroot(x)) { ull y=fa[x],z=fa[y]; if (!isroot(y)) { if (c[y][0]==x^c[z][0]==y) rotata(x); else rotata(y); } rotata(x); } } void access(ull x) { ull t(0); while (x) { splay(x); c[x][1]=t; t=x; updata(x); x=fa[x]; } } void makeroot(ull x) { access(x); splay(x); rev[x]^=1; } void link(ull x,ull y) { makeroot(x); fa[x]=y; } void cut(ull x,ull y) { makeroot(x); access(y); splay(y); c[y][0]=fa[x]=0; } void split(ull x,ull y) { makeroot(y); access(x); splay(x); } int main() { while(scanf("%llu",&n)!=EOF) { for (ull i=1; i<=n; i++) v[i]=0,size[i]=s[i]=mt[i]=1; memset(c,0,sizeof c); memset(fa,0,sizeof fa); for (ull i=2; i<=n; i++) { scanf("%llu",&y); link(y,i); } scanf("%llu",&q); for (ull i=1; i<=q; i++) { scanf("%llu%llu%llu",&op,&x,&y); if (op==2)//+w { scanf("%llu",&w); split(x,y); paint(x,1,w); } if (op==3)//not 2^64-1 18446744073709551615 { split(x,y); paint(x,-1,18446744073709551615ull); } if (op==1)//*w { scanf("%llu",&w); split(x,y); paint(x,w,0); } if (op==4)//u,v { split(x,y); printf("%llu\n",s[x]); } } } return 0; }
F. Modular Production Line
An automobile factory has a car production line. Now the market is oversupply and the production line is often shut down. To make full use of resources, the manager divides the entire production line into NN parts (1...N)(1...N). Some continuous parts can produce sub-products. And each of sub-products has their own value. The manager will use spare time to produce sub-products to make money. Because of the limited spare time, each part of the production line could only work at most KK times. And Because of the limited materials, each of the sub-products could be produced only once. The manager wants to know the maximum value could he make by produce sub-products.
Input
The first line of input is TT, the number of test case.
The first line of each test case contains three integers, N, KN,K and MM. (MM is the number of different sub-product).
The next MM lines each contain three integers A_i, B_i, W_iAi,Bi,Wi describing a sub-product. The sub-product has value W_iWi. Only A_iAi to B_iBi parts work simultaneously will the sub-product be produced (include A_iAi to B_iBi).
1 \le T \le 1001≤T≤100
1 \le K \le M \le 2001≤K≤M≤200
1 \le N \le 10^51≤N≤105
1 \le A_i \le B_i \le N1≤Ai≤Bi≤N
1 \le W_i \le 10^51≤Wi≤105
Output
For each test case output the maximum value in a separate line.
样例输入复制
4 10 1 3 1 2 2 2 3 4 3 4 8 10 1 3 1 3 2 2 3 4 3 4 8 100000 1 3 1 100000 100000 1 2 3 100 200 300 100000 2 3 1 100000 100000 1 150 301 100 200 300
样例输出复制
10 8 100000 100301
题目来源
建图网络流
先对区间的端点进行离散化，然后建边
#include<bits/stdc++.h> using namespace std; const int N=1e4+5; const int M=1e5+5; const int INF=0x3f3f3f3f; int FIR[N],TO[M],CAP[M],FLOW[M],COST[M],NEXT[M],tote; int pre[N],dist[N],q[400000]; bool vis[N]; int n,m,S,T; void init() { tote=0; memset(FIR,-1,sizeof(FIR)); } void add(int u,int v,int cap,int cost) { TO[tote]=v; CAP[tote]=cap; FLOW[tote]=0; COST[tote]=cost; NEXT[tote]=FIR[u]; FIR[u]=tote++; TO[tote]=u; CAP[tote]=0; FLOW[tote]=0; COST[tote]=-cost; NEXT[tote]=FIR[v]; FIR[v]=tote++; } bool SPFA(int s, int t) { memset(dist,INF,sizeof(dist)); memset(vis,false,sizeof(vis)); memset(pre,-1,sizeof(pre)); dist[s] = 0; vis[s]=true; q[1]=s; int head=0,tail=1; while(head!=tail) { int u=q[++head]; vis[u]=false; for(int v=FIR[u]; v!=-1; v=NEXT[v]) { if(dist[TO[v]]>dist[u]+COST[v]&&CAP[v]>FLOW[v]) { dist[TO[v]]=dist[u]+COST[v]; pre[TO[v]]=v; if(!vis[TO[v]]) { vis[TO[v]] = true; q[++tail]=TO[v]; } } } } return pre[t]!=-1; } void MCMF(int s, int t, int &cost, int &flow) { flow=cost=0; while(SPFA(s,t)) { int Min=INF; for(int v=pre[t]; v!=-1; v=pre[TO[v^1]]) Min=min(Min, CAP[v]-FLOW[v]); for(int v=pre[t]; v!=-1; v=pre[TO[v^1]]) { FLOW[v]+=Min; FLOW[v^1]-=Min; cost+=COST[v]*Min; } flow+=Min; } } int l[205],r[205],c[205]; int a[405],b[100005]; int main() { int ca,k; scanf("%d",&ca); while(ca--) { scanf("%d%d%d",&n,&k,&m); init(); int num=0; for(int i=1; i<=m; i++) { scanf("%d%d%d",&l[i],&r[i],&c[i]),r[i]+=1; a[++num]=l[i],a[++num]=r[i]; } sort(a+1,a+num+1); num=unique(a+1,a+num+1)-a-1; for(int i=1;i<=num;i++)b[a[i]]=i; S=0; T=num+1; for(int i=0;i<=num;i++)add(i,i+1,k,0); for(int i=1;i<=m;i++)add(b[l[i]],b[r[i]],1,-c[i]); int cost,flow; MCMF(S,T,cost,flow); printf("%d\n",-cost); } return 0; }
G. Give Candies
There are NN children in kindergarten. Miss Li bought them NN candies. To make the process more interesting, Miss Li comes up with the rule: All the children line up according to their student number (1...N)(1...N), and each time a child is invited, Miss Li randomly gives him some candies (at least one). The process goes on until there is no candy. Miss Li wants to know how many possible different distribution results are there.
Input
The first line contains an integer TT, the number of test case.
The next TT lines, each contains an integer NN.
1 \le T \le 1001≤T≤100
1 \le N \le 10^{100000}1≤N≤10100000
Output
For each test case output the number of possible results (mod 1000000007).
样例输入复制
1 4
样例输出复制
8
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