kattis Curious Cupid (莫队算法)

时间：2017-02-27 22:54:57 阅读：222 评论：0 收藏：0 [点我收藏+]

标签：sam output tom inpu ++ hidden memset back 莫队

Curious Cupid

There are $K$

different languages in the world. Each person speaks one and only one language. There are exactly $N$

single women.

Cupid, the god of love, wants to match every single man to a single woman, and vice versa. Everybody wants to find a partner who speaks the same language as s/he does. Communication between the couple is very important! Cupid asks these $N$

$N$

It is too hard for Cupid to match all people at the same time. What Cupid does is to repeatedly look at some specific interval in these two lines, pick the men and women in that interval and find the maximum number of man-woman pairs who speak the same language and can be matched.

Input

The first line contains three integers $N$

$N$

.
The second line contains $N$

integers

a_{0}, a_{1}, a_{2}, \dots, a_{N - 1}

th man.
The third line contains $N$

a_{0}, a_{1}, a_{2}, \dots, a_{N - 1}

th woman.
In the next $M$

a_{0}, a_{1}, a_{2}, \dots, a_{N - 1}

Output

For each interval, print the maximum number of couples Cupid could match.

Sample Input 1	Sample Output 1
3 4 2 0 0 1 0 0 0 0 0 2 2 0 1 1 2	1 0 2 1

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
typedef long long ll;
using namespace std;
const int N = 5e4+10;
const int M = 1e6+10;
ll num[N],up[N],dw[N],ans,aa,bb,cc;
int n,m,k;
int x[N],y[N],pos[N];
int cntx[M],cnty[M];
struct qnode {
    int l,r,id;
} qu[N];
bool cmp(qnode a,qnode b) {
    if(pos[a.l]==pos[b.l])
        return a.r<b.r;
    return pos[a.l]<pos[b.l];
}
void update(int p,int d) {
    if(x[p]==y[p]) {
        cntx[x[p]]+=d;
        cnty[x[p]]+=d;
        ans+=d;
    } else {
        if(d==1) {
            if(cnty[x[p]]>cntx[x[p]]) {
                ans+=d;
            }
            if(cntx[y[p]]>cnty[y[p]]) {
                ans+=d;
            }
        } else {
            if(cnty[x[p]]>=cntx[x[p]]) {
                ans+=d;
            }
            if(cntx[y[p]]>=cnty[y[p]]) {
                ans+=d;
            }
        }
        cntx[x[p]]+=d;
        cnty[y[p]]+=d;
    }
}
int main() {
    int bk,pl,pr,id;
    scanf("%d%d%d",&n,&m,&k);
    memset(num,0,sizeof num);
    bk=ceil(sqrt(1.0*n));
    for(int i=1; i<=n; i++) {
        scanf("%d",&x[i]);
        pos[i]=(i-1)/bk;
    }
    for(int i=1; i<=n; i++) {
        scanf("%d",&y[i]);
    }
    for(int i=0; i<m; i++) {
        scanf("%d%d",&qu[i].l,&qu[i].r);
        qu[i].l++;
        qu[i].r++;
        qu[i].id=i;
    }
    sort(qu,qu+m,cmp);
    pl=1,pr=0;
    ans=0;
    for(int i=0; i<m; i++) {
        id=qu[i].id;
        if(pr<qu[i].r) {
            for(int j=pr+1; j<=qu[i].r; j++)
                update(j,1);
        } else {
            for(int j=pr; j>qu[i].r; j--)
                update(j,-1);
        }
        pr=qu[i].r;
        if(pl<qu[i].l) {
            for(int j=pl; j<qu[i].l; j++)
                update(j,-1);
        } else {
            for(int j=pl-1; j>=qu[i].l; j--)
                update(j,1);
        }
        pl=qu[i].l;
        up[id]=ans;
    }

    for(int i=0; i<m; i++)
        printf("%d\n",up[i]);
    return 0;
}

kattis Curious Cupid (莫队算法)

标签：sam output tom inpu ++ hidden memset back 莫队

原文地址：http://www.cnblogs.com/jianrenfang/p/6476752.html

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