4443: [Scoi2015]小秃玩矩阵二分答案+最大流-优快云博客

本文介绍了一个结合二分查找与Dinic最大流算法解决特定问题的例子。该问题涉及通过调整阈值来确定最优解，利用Dinic算法求解网络流问题，并通过二分查找来逼近最佳阈值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

签到题

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 505
#define M 600005
#define inf 1000000007
using namespace std;
int n,m,k,T,cnt;
int a[255][255];
int head[N],dis[N],q[N],cur[N];
int next[M],list[M],key[M];
inline int read()
{
    int a=0,f=1; char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
    while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}
    return a*f;
}
inline void insert(int x,int y,int z)
{
    next[++cnt]=head[x];
    head[x]=cnt;
    list[cnt]=y;
    key[cnt]=z;
}
inline bool BFS()
{
    memset(dis,-1,sizeof(dis));
    int t=0,w=1,x;
    dis[0]=1; q[1]=0;
    while (t<w)
    {
        x=q[++t];
        for (int i=head[x];i;i=next[i])
            if (key[i]&&dis[list[i]]==-1)
                dis[q[++w]=list[i]]=dis[x]+1;
    }
    return dis[T]!=-1;
}
int find(int x,int flow)
{
    if (x==T) return flow;
    int w,used=0;
    for (int i=cur[x];i;i=next[i])
        if (key[i]&&dis[list[i]]==dis[x]+1)
        {
            w=find(list[i],min(key[i],flow-used));
            key[i]-=w; key[i^1]+=w; used+=w;
            if (used==flow) return used;
        }
    if (!used) dis[x]=-1;
    return used;
}
inline int dinic()
{
    int tmp=0;
    while (BFS())
    {
        for (int i=0;i<=T;i++) cur[i]=head[i];
        tmp+=find(0,inf);
    }
    return tmp;
}
inline void build(int mid)
{
    memset(head,0,sizeof(head)); 
    cnt=1;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            if (a[i][j]<mid)
                insert(i,j+n,1),insert(j+n,i,0);
    for (int i=1;i<=n;i++)
        insert(0,i,1),insert(i,0,0);
    for (int i=1;i<=m;i++)
        insert(i+n,T,1),insert(T,i+n,0);
}
int main()
{
    n=read(); m=read(); k=n+1-read(); T=n+m+1;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++) 
            a[i][j]=read();
    int l=1,r=1e9;
    while (l<=r)
    {
        int mid=l+r>>1;
        build(mid);
        if (dinic()>k-1) r=mid-1;
        else l=mid+1;
    }
    cout << r << endl;
}