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Solution

看到什麽最大值最小肯定二分啊。
check直接跑一個二分圖匹配就好了。
orz ztl！！！

代碼實現

/*
  mail: [email protected]
  author: MLEAutoMaton
  This Code is made by MLEAutoMaton
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define ll long long
#define re register
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
inline int gi()
{
    int f=1,sum=0;char ch=getchar();
    while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
    return f*sum;
}
const int N=500010,M=2000010,Inf=1e9+10;
struct node
{
    int to,nxt,w;
}e[M<<1];
int front[N],cnt,s,t,dep[N],n,m,k,a[510][510];
void Add(int u,int v,int w)
{
    e[cnt]=(node){v,front[u],w};front[u]=cnt++;
    e[cnt]=(node){u,front[v],0};front[v]=cnt++;
}
void clear(){memset(front,-1,sizeof(front));cnt=0;}
queue<int>Q;
bool bfs()
{
    Q.push(s);memset(dep,0,sizeof(dep));
    dep[s]=1;
    while(!Q.empty())
    {
        int u=Q.front();Q.pop();
        for(int i=front[u];~i;i=e[i].nxt)
        {
            int v=e[i].to;
            if(e[i].w && !dep[v])
            {
                dep[v]=dep[u]+1;Q.push(v);
            }
        }
    }
    return dep[t];
}
int dfs(int u,int flow)
{
    if(u==t || !flow)return flow;
    for(int i=front[u];~i;i=e[i].nxt)
    {
        int v=e[i].to;
        if(e[i].w && dep[v]==dep[u]+1)
        {
            int di=dfs(v,min(flow,e[i].w));
            if(di)
            {
                e[i].w-=di;e[i^1].w+=di;
                return di;
            }
            else dep[v]=0;
        }
    }
    return 0;
}
int dinic()
{
    int flow=0;
    while(bfs())
        while(int d=dfs(s,Inf))flow+=d;
    return flow;
}
void build(int mid)
{
    for(int i=1;i<=n;i++)
        Add(s,i,1);
    for(int i=1;i<=m;i++)
        Add(i+n,t,1);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            if(a[i][j]<=mid)Add(i,j+n,1);
}
int main()
{
    n=gi();m=gi();k=gi();
    clear();int Max=0;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        {
            a[i][j]=gi();
            Max=max(Max,a[i][j]);
        }
    int l=0,r=Max,ans=0;t=n+m+1;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        clear();
        build(mid);
        if(dinic()>=n-k+1){r=mid-1;ans=mid;}
        else l=mid+1;
    }
    printf("%d\n",ans);
    return 0;
}
 

【洛谷4251】 [SCOI2015]小凸玩矩陣（二分答案，二分圖匹配）