[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=1001

[算法]

最小割

[代碼]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1010
const long long inf = 1e18;

struct edge
{
        int to;
        long long w;
        int nxt;
} e[MAXN * MAXN * 8];

int i,j,n,m,tot,S,T,x;
 
int head[MAXN * MAXN],depth[MAXN * MAXN];
long long ans,flow;

template <typename T> inline void read(T &x)
{
    long long f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) 
        { 
                if (c == ‘-‘) f = -f; 
        }
    for (; isdigit(c); c = getchar()) x = (x << 3
 
) + (x << 1) + c - ‘0‘;
    x *= f;
}
inline int getid(int x,int y)
{
        return (x - 1) * m + y;
}
inline void addedge(int u,int v,long long w)
{
        tot++;
        e[tot] = (edge){v,w,head[u]};
        head[u] = tot;
        tot++;
        e[tot] = (edge){u,w,head[v]};
        head[v] 
 
= tot;
} 
inline bool bfs()
{
        int i,u,v,w,l,r;
        static int q[MAXN * MAXN];
        for (i = 1; i <= n * m; i++) depth[i] = 0;
        q[l = r = 1] = S;
        depth[S] = 1;
        while (l <= r)
        {
                u = q[l];
                l++;
                for (i = head[u]; i; i = e[i].nxt)
                {
                        v = e[i].to;
                        w = e[i].w;    
                        if (!depth[v] && w)
                        {
                                depth[v] = depth[u] + 1;
                                q[++r] = v;
                                if (v == T) return true;
                        }
                }        
        }    
        return false;
}
inline long long dinic(int u,long long flow)
{
        int i,v;
        long long w,rest = flow,k;
        if (u == T) return flow;
        for (i = head[u]; i && rest; i = e[i].nxt)
        {
                v = e[i].to;
                w = e[i].w;
                if (depth[v] == depth[u] + 1 && w)
                {
                        k = dinic(v,min(rest,w));
                        if (!k) depth[v] = 0;
                        e[i].w -= k;
                        e[i ^ 1].w += k;
                        rest -= k;
                }
        }
        return flow - rest;
}
int main() 
{
        
        read(n); read(m);
        tot = 1;
        for (i = 1; i <= n; i++)
        {
                for (j = 1; j < m; j++)
                {
                        read(x);
                        addedge(getid(i,j),getid(i,j + 1),x);
                }
        }
        for (i = 1; i < n; i++)
        {
                for (j = 1; j <= m; j++)
                {
                        read(x);
                        addedge(getid(i,j),getid(i + 1,j),x);
                }
        }
        for (i = 1; i < n; i++)
        {
                for (j = 1; j < m; j++)
                {
                        read(x);
                        addedge(getid(i,j),getid(i + 1,j + 1),x);
                }
        }
        ans = 0; 
        S = 1; T = n * m;
        while (bfs()) 
        {
                while (flow = dinic(S,inf)) ans += flow;
        }
        printf("%lld\n",ans);
        
        return 0;
    
}

[BZOJ 2006] 狼抓兔子