2016 UESTC Training for Data Structures(3)

J 郭大侠与Rabi-Ribi
超级水题
按时间排个序在扫一道OK

#include<queue>
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
 int ans,heat,n;
priority_queue<int>q;
struct s 
{
int a;
int b;
}f[100005];
bool cmp(s aa,s bb)
{
    return aa.a>bb.a;
}
int main()
{
    cin>>n;
    for (int i=1;i<=n;i++)
    scanf("%d",&f[i].a);
    for (int i=1;i<=n;i++)
    scanf("%d",&f[i].b);
    sort(f+1,f+n+1,cmp);
    heat=1;
    for(int i=f[1].a;i>0;i--)
    {
    while (i==f[heat].a)
    q.push(f[heat++].b);
    if (!q.empty())
    {
        ans+=q.top();
        q.pop();
    } 
    }
    cout<<ans<<endl;
    return 0;
}

K 郭大侠与甲铁城
神奇的莫队算法,轻易莽掉了这一题

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
struct node{
    int l,r,wei;
}e[100005];
int un,n,q,a[100005],ans,cnt[100005],w[100005],aa;
bool cmp(node a,node b)
{
    int i=a.l/un;int j=b.l/un;
    if (i!=j) return i<j;
    else return a.r<b.r;
}
int main()
{
    scanf("%d%d",&n,&q);
    un=sqrt(n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        e[i].wei=i;
    }
    for (int i=1;i<=q;i++)
    {
        scanf("%d%d",&e[i].l,&e[i].r);  
    }
    int l=0,r=-1;ans=0;
    sort(e+1,e+1+q,cmp);
    for (int i=1;i<=q;i++)
    {
        while(r<e[i].r)
        {
            r++;
            if (cnt[a[r]]==0)
            {
                ans++;
            }
            cnt[a[r]]++;
        }
        while (r>e[i].r)
        {
            if (cnt[a[r]]==1) ans--;
            cnt[a[r]]--;
            r--;
        }
        while(l<e[i].l)
        {
            if (cnt[a[l]]==1)
            {
                ans--;
            }
            cnt[a[l]]--;    
            l++;
        } 
        while(l>e[i].l)
        {
            l--;
            if (cnt[a[l]]==0)
            {
                ans++;
            }
            cnt[a[l]]++;
        }
        w[e[i].wei]=ans;
    }
    for (int i=1;i<=q;i++)
    printf("%d\n",w[i]);
}

L 郭大侠与苦恼
题意:给了你一棵树(节点>100000),求从u点到v点的简单路径上点编号的异或和等于0的(u,v)有多少
可以想到求d[x]^d[y]^lca(x,v)==0
用map的启发式合并
什么是启发式合并?并查集就是最经典的一种.
其实就是用map维护dfs序,然而要MLE,所以就用了启发式合并,就是把u往v上合并.

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
struct node{
    int v,next;
}e[200005];
int head[100005],k=1,fa[100005],n,x,y,aa;
long long ans;
map <int,int> s[100005];
map <int,int> p[100005];
void add(int u,int v)
{
    e[k].v=v;
    e[k].next=head[u];
    head[u]=k++;
}
void unio(map<int,int > &a,map<int,int> &b)
{
    if (a.size()>b.size())
    {
        swap(a,b);
    }   
    map <int,int> :: iterator it=a.begin();
    while (a.size())
    {
        it=a.begin();
        b[it->first]+=it->second;
        a.erase(it);
    }
}
void dfs(int x,int dep)
{
    p[x][dep]++;
    for (int i=head[x];i;i=e[i].next)
    {
        int v=e[i].v;
        if (v==fa[x]) continue;
        fa[v]=x;
        dfs(v,dep^v);
        map <int,int> :: iterator it=p[v].begin();
        while (it!=p[v].end())
        {   
            int S = ((it->first)^x);
            if(p[x].count(S)) ans += 1LL * p[x][S] * it->second;
            it++;
        }
        unio(p[v],p[x]);
    }
}

int main()
{
    scanf("%d",&n);
    for (int i=1;i<n;i++)
    {
        scanf("%d%d",&x,&y);
        add(x,y);
        add(y,x);
    }
    for (int i=1;i<=n;i++)
    {
        p[i]=s[i];
    }
    dfs(1,1);
    printf("%lld\n",ans);
}

M 卿学姐失恋了Ⅱ
用个dfsok,水的惊心动魄

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m,a[25],hanoi[25],sd;
long long work(int g,int w)
{
    if (g==0) return 0;
    while (a[g]==w&&g>0) 
    g--;
    if (!g) return 0;
    long long res=work(g-1,6-a[g]-w);
    return res+hanoi[g-1];
}
int main()
{
    scanf("%d",&m);
    for (int i=20;i>=1;i--)
    {
        scanf("%d",&a[i]);
    }
    hanoi[0]=1;
    for (int i=1;i<=20;i++)
    {
        hanoi[i]=hanoi[i-1]*2;
    }
    if (work(20,1)<=m) printf("YES\n");
    else if (work(20,2)<=m) printf("YES\n");
    else if (work(20,3)<=m) printf("YES\n");
    else printf("NO\n");
}

N 秋实大哥搞算数
暴力模拟即可

#include <cstdio>
#include <cmath>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
long long f[1000005],t,sdfr;
char s[1000005];
int main()
{
    scanf("%d\n",&t);
    while(t--)
    {
        scanf("%s",s);
        int l=strlen(s);
        int x=0;
        int zf=1;
        int ge=0;
        int fuhao=0;
        for (int i=0;i<=l;i++)
        {
            char ch=s[i];
            if (ch>='0'&&ch<='9')
            {
                x=x*10+(ch-'0');
            } 
            else 
            {   
                f[++ge]=x*zf;   
                zf=1;
                x=0;
                if (fuhao==1)
                {
                    f[ge-1]=f[ge-1]*f[ge];
                    ge--;
                }
                if (fuhao==3)
                {
                    f[ge-1]=f[ge-1]/f[ge];
                    ge--;
                }
                fuhao=0;
                if (ch=='*') fuhao=1;
                if (ch=='/') fuhao=3;
                if (ch=='-') zf=-1;
            }
        }
        f[++ge]=x*zf;
        if (fuhao==1)
            {
                f[ge-2]=f[ge-1]*f[ge-2];
                ge--;
            }
            if (fuhao==3)
            {
                f[ge-2]=f[ge-1]/f[ge-2];
                ge--;
            }
        long long ans=0;
        for (int i=1;i<ge;i++)
        {
            ans+=f[i];
        }
        printf("%lld\n",ans);
    }
    return 0;
}

O 卿学姐种美丽的花
线段树

#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <algorithm>  
#include <queue>  
using namespace std;  
#define mod 772235  
#define ll long long  
#define maxn 1000005  
#define lrt (rt<<1)  
#define rrt ((rt<<1)|1)  
long long a[maxn];  
int N, Q;  
struct node  
{  
    int l, r;  
    ll lazy, cnt;  
}tr[maxn * 4];  
void bulid(int rt, int l, int r)  
{  
    tr[rt].l = l; tr[rt].r = r;  
    if (l == r)  
    {  
        return;  
    }  
    int mid = (l + r) >> 1;  
    bulid(lrt, l, mid);  
    bulid(rrt, mid + 1, r);  
}  
void push_down(int rt)  
{  
    if (tr[rt].lazy != 0)  
    {  
        tr[lrt].cnt += tr[rt].cnt, tr[rrt].cnt += tr[rt].cnt;  
        int mid = tr[lrt].r;  
        ll v = tr[rt].lazy; 
        tr[lrt].lazy += v;  
        v = tr[rt].lazy - (mid + 1 - tr[rt].l)*tr[rt].cnt;  
        tr[rrt].lazy += v;  
        tr[rt].lazy = 0;    
        tr[rt].cnt = 0;  
    }  
}  
void update(int rt, int l, int r, long long v)  
{  
    if (l == tr[rt].l&&tr[rt].r == r)  
    {  
        tr[rt].lazy += v;  
        tr[rt].cnt++;  
        return;  
    }  
    push_down(rt);  
    int mid = (tr[rt].l + tr[rt].r) >> 1;  
    if (r <= mid) update(lrt, l, r, v);  
    else if (l > mid) update(rrt, l, r, v);  
    else  
    {  
        update(lrt, l, mid, v);  
        update(rrt, mid + 1, r, v - (mid + 1 - l));  
    }  
}  
long long query(int rt, int l, int r)  
{  
    if (l == tr[rt].l&&r == tr[rt].r) return tr[rt].lazy;  
    push_down(rt);  
    int mid = (tr[rt].l + tr[rt].r) >> 1;  
    if (r <= mid) return query(lrt, l, r);  
    else if (l > mid) return query(rrt, l, r);  
    else return query(lrt, l, mid) + query(rrt, mid + 1, r);  
}  
int main()  
{  
    scanf("%d%d", &N, &Q);  
    for (int i = 1; i <= N; ++i)  
        scanf("%lld", &a[i]);  
    bulid(1, 1, N);  
    int t, x, y, l, r;  
    ll ans;  
    for (int i = 0; i < Q; ++i)  
    {  
        scanf("%d", &t);  
        if (t == 1)  
        {  
            scanf("%d%d", &x, &y);  
            l = x;  
            r = min(x + y - 1, N);  
            update(1, l, r, y);    
        }  
        else if (t == 2)  
        {  
            scanf("%d", &x);  
            ans = (query(1, x, x)+a[x]) % mod;  
            printf("%lld\n", ans);  
        }  
    }  
    return 0;  
}