P3033 [USACO11NOV]牛的障碍Cow Steeplechase

P3033 [USACO11NOV]牛的障碍Cow Steeplechase

套路套路全是套路！二分图匹配问题全是套路！

这里分析一类常见的二分图匹配：线段相交

给定n根平行与x或y轴的线段，求最大独立集

最大独立集=总数-最小割=总数-最大流

匹配一下即可

记住！ij不分见祖宗！x1>x2,y1>y2未判见祖宗！

代码：

#include<bits/stdc++.h>
using namespace std;
#define re register 
#define il inline 

const int N=505;
const int M=500005;
struct line{
    int x1,y1,x2,y2;
}K[N],K1[N],K2[N];
int cnt1=0;
int cnt2=0;
struct node{
    int x,y;
    int val;
};
int n;
int s[N][N];
int l=1,r,ans;
node a[N*N+100];
int c[N][N];
int sx[N],sy[N],us[N]; 
il bool cmp(node x,node y){
    return x.val<y.val;
}
il bool work(int u){
    //cout<<u<<endl;
    for(re int v=1;v<=cnt2;v++){
        if(c[u][v]&&!us[v]){
            us[v]=1;
            if(sy[v]==-1||work(sy[v])){
                sx[u]=v;
                sy[v]=u;
                return 1;
            }
        }
    } 
    return 0;
}
il int check(){
    int sum=0;
    memset(c,0,sizeof(c));
    memset(sx,-1,sizeof(sx));
    memset(sy,-1,sizeof(sy));
    for(re int i=1;i<=cnt1;i++){
        for(re int j=1;j<=cnt2;j++){
            if(K1[i].x1>=K2[j].x1&&K1[i].x1<=K2[j].x2&&K2[j].y1>=K1[i].y1&&K2[j].y1<=K1[i].y2){
                c[i][j]=1;
            }
        }
    }
    for(int i=1;i<=cnt1;i++){
        memset(us,0,sizeof(us));
        sum+=work(i);
    }
    return sum;
}
int main(){
    scanf("%d",&n);
    for(re int i=1;i<=n;i++){
        scanf("%d%d%d%d",&K[i].x1,&K[i].y1,&K[i].x2,&K[i].y2);
        if(K[i].x1>K[i].x2) swap(K[i].x1,K[i].x2);
        if(K[i].y1>K[i].y2) swap(K[i].y1,K[i].y2);
        if(K[i].x1==K[i].x2) K1[++cnt1]=K[i];
        else K2[++cnt2]=K[i]; 
    }
    printf("%d\n",n-check());
    return 0;
} 

 

转载于:https://www.cnblogs.com/QYJ060604/p/11520632.html