hdu3041

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链接:hdu3041

Description

  • Bessie wants to navigate her spaceship through a dangerous asteroid field in the shape of an N x N grid (1 <=
  • N <= 500). The grid contains K asteroids (1 <= K <= 10,000), which are conveniently located at the lattice

  • points of the grid.

  • Fortunately, Bessie has a powerful weapon that can vaporize all the asteroids in any given row or column of

  • the grid with a single shot.This weapon is quite expensive, so she wishes to use it sparingly.Given the
  • location of all the asteroids in the field, find the minimum number of shots Bessie needs to fire to
  • eliminate all of the asteroids.

    Input

  • Line 1: Two integers N and K, separated by a single space.
  • Lines 2..K+1: Each line contains two space-separated integers R and C (1 <= R, C <= N) denoting the row and
  • column coordinates of an asteroid, respectively.

    Output

  • Line 1: The integer representing the minimum number of times Bessie must shoot.

    Sample Input

    3 4
    1 1
    1 3
    2 2
    3 2

    Sample Output

    2

    题解

  • 二分图最小点覆盖
  • 其实就等于二分图最大匹配
  • 数组忘了清零，WA一发，数组开大十倍，TLE一发
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#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,x,y) for(register int i=x;i<=y;++i)
#define repd(i,x,y) for(register int i=x;i>=y;--i)
#define ll long long
using namespace std;
const int N=1e5+7;
struct node{
    int v,nxt;
    inline void init(int a,int b){v=a; nxt=b;}
}e[N];
int head[N],cnt=1,vis[N],mark[N],n,m,ans;
template <typename T>inline void read(T &x){
    char c;int sign=1;x=0;
    do{c=getchar(); if(c=='-')sign=-1;}while(c<'0'||c>'9');
    do{x=x*10+c-'0'; c=getchar();}while(c>='0'&&c<='9');
    x*=sign;
}

void reade(int a,int b){e[cnt].init(b,head[a]); head[a]=cnt++;}

int find(int u){
    for(int i=head[u];i;i=e[i].nxt)if(!vis[e[i].v]){
        vis[e[i].v]=1;
        if(!mark[e[i].v]||find(mark[e[i].v])){
            mark[e[i].v]=u;
            return 1;
        }
    }
    return 0;
}

int main(){
    while(~scanf("%d%d",&n,&m)){
        ans=0;
        memset(head,0,sizeof(head));
        memset(mark,0,sizeof(mark));
        rep(i,1,m){
            int x,y;
            read(x);read(y);
            reade(x,y);
        }
        rep(i,1,n){
            memset(vis,0,sizeof(vis));
            ans+=find(i);
        }
        printf("%d\n",ans);
    }
    return 0;
}