[bzoj4443][Scoi2015]小凸玩矩陣（二分答案+二分圖最大匹配）

【題目鏈接】http://www.lydsy.com/JudgeOnline/problem.php?id=4443
【解題思路】二分答案轉化爲判定性問題，將數值小於等於二分值的節點拿去建二分圖，跑一邊最大匹配即可
【呆馬】

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<iostream>
const int N=251;
using namespace std;
struct st{int to,next;} e[N*N];
int n,m,k,i,j,t,cnt,l,r,mid,ans,a[N][N],b[N*N],fi[N],f[N];
bool vis[N];
void add(int x,int y){e[++cnt].to=y; e[cnt].next=fi[x]; fi[x]=cnt;}
bool find(int x)
{
    for (int i=fi[x];i;i=e[i].next)
        if (!vis[e[i].to])
        {
            int y=e[i].to;
            vis[y]=1;
            if (!f[y] || find(f[y])){f[y]=x; return 1;}
        }
    return 0;
}

bool check(int x)
{
    int num=0;
    cnt=0;
    for (int i=1;i<=n;i++) fi[i]=0;
    for (int i=1;i<=m;i++) f[i]=0;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            if (a[i][j]<=x) add(i,j);
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=m;j++) vis[j]=0;
        if (find(i)) num++;
        if (num==k) return 1;
    }
    return 0;
}

int main()
{
        scanf("%d%d%d\n",&n,&m,&k);
        k=n-k+1;
        for (i=1;i<=n;i++,scanf("\n"))
            for (j=1;j<=m;j++)
            {
                scanf("%d",&a[i][j]);
                b[++t]=a[i][j];
            }
        sort(b+1,b+t+1);
        t=unique(b+1,b+t+1)-(b+1);
        for (l=1,r=t;l<=r;)
        {
            mid=(l+r)>>1;
            if (check(b[mid])) ans=b[mid],r=mid-1;
            else l=mid+1;
        }
        printf("%d",ans);
}