【洛谷4001】 [ICPC-Beijing 2006]狼抓兔子

传送门

洛谷

Solution

直接跑最小割板子就好了。

代码实现

#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=1000010,NN=1001,Inf=1e9+10;
int id[NN][NN],front[N],cnt,cur[N];
int dep[N],s,t,n,m,tot,sum;
struct node
{
    int to,nxt,w;
}e[12000000];
void Add(int u,int v,int w)
{
    e[cnt]=(node){v,front[u],w};front[u]=cnt++;
    e[cnt]=(node){u,front[v],w};front[v]=cnt++;
}
queue<int>Q;
bool bfs()
{
    memset(dep,0,sizeof(dep));
    dep[s]=1;
    Q.push(s);
    while(!Q.empty())
    {
        int u=Q.front();Q.pop();
        for(int i=front[u];i!=-1;i=e[i].nxt)
        {
            int v=e[i].to;
            if(!dep[v] && e[i].w)
            {
                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=cur[u];i!=-1;i=e[i].nxt)
    {
        int v=e[i].to;
        if(dep[v]==dep[u]+1 && e[i].w)
        {
            int di=dfs(v,min(e[i].w,flow));
            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())
    {
        for(int i=s;i<=t;i++)cur[i]=front[i];
        while(int d=dfs(s,Inf))flow+=d;
    }
    return flow;
}
int main()
{
    n=gi();m=gi();
    memset(front,-1,sizeof(front));
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)id[i][j]=++tot;
    t=tot++;
    Add(s,id[1][1],Inf);
    Add(id[n][m],t,Inf);
    for(int i=1;i<=n;i++)
        for(int j=1;j<m;j++)
        {
            int x=gi();
            Add(id[i][j],id[i][j+1],x);
            Add(id[i][j+1],id[i][j],x);
        }
    for(int i=1;i<n;i++)
        for(int j=1;j<=m;j++)
        {
            int x=gi();
            Add(id[i][j],id[i+1][j],x);
            Add(id[i+1][j],id[i][j],x);
        }
    for(int i=1;i<n;i++)
        for(int j=1;j<m;j++)
        {
            int x=gi();
            Add(id[i][j],id[i+1][j+1],x);
            Add(id[i+1][j+1],id[i][j],x);
        }
    printf("%d\n",Dinic()/2);
    return 0;
}

标签：Beijing,洛谷,di,int,4001,flow,dep,return,include
来源： https://www.cnblogs.com/mle-world/p/10555689.html