bzoj2132

最小割

套路最小割。。。

盗一波图 来自GXZ神犇

对于这样的图，我们要么割ai,bj,要么割bi,aj,要么割ai,ci+cj,aj,要么割bi,ci+cj,bj,然后这样建图跑最小割就行了

但这不是重点，这道题我t了大概一个月，不知道为什么，怎么和别人比对代码好像没有什么差异，结果发现判断delta=0不能放在for循环里，否则会很慢。。。俞勇的红书不靠谱啊。。。怪不得我的网络流那么慢。。。

 
#include<bits/stdc++.h>
using namespace std;
const int N = 10010, inf = 1000000010;
const int dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
int head[N], d[N], q[N], iter[N];
struct edge {
    int nxt, to, f;
} e[N * 50];
int n, cnt = 1, source, sink, ans, m; 
#define id(i, j) (i - 1) * m + j
int read()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}
void link(int u, int v, int f)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].f = f;
}
void insert(int u, int v, int f)
{
    link(u, v, f);
    link(v, u, 0);
}
bool bfs()
{
    queue<int> q;
    q.push(source);
    memset(d, 0, sizeof(d));
    d[source] = 1;
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i; i = e[i].nxt) if(e[i].f && !d[e[i].to])
        {
            d[e[i].to] = d[u] + 1;
            q.push(e[i].to);
            if(e[i].to == sink) return true;
        }
    }
    return false;
}
int dfs(int u, int delta)
{
    if(u == sink || delta == 0) return delta;
    int ret = 0;
    for(int &i = iter[u]; i; i = e[i].nxt) if(e[i].f && d[e[i].to] == d[u] + 1)
    {
        int x = dfs(e[i].to, min(e[i].f, delta));
        if(x == 0) d[e[i].to] = 0;
        e[i].f -= x;
        e[i ^ 1].f += x;
        delta -= x;
        ret += x;
        if(delta == 0) return ret;
    }
    return ret;
}
int dinic()
{
    int ret = 0;
    while(bfs())
    {
        for(int i = source; i <= sink; ++i) iter[i] = head[i];
        ret += dfs(source, inf); 
    }
    return ret;
}
int main()
{
    scanf("%d%d", &n, &m);
    sink = n * m + 1; 
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
        {
            int x, a = id(i, j);
            scanf("%d", &x);
            ans += x;
            if((i + j) & 1) insert(source, a, x);
            else insert(a, sink, x);
        }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
        {
            int x, a = id(i, j);
            scanf("%d", &x);
            ans += x;
            if((i + j) & 1) insert(a, sink, x);
            else insert(source, a, x);
        }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j)
        {
            int x, a = id(i, j), b;
            scanf("%d", &x);
            for(int k = 0; k < 4; ++k)
            {
                int xx = i + dx[k], yy = j + dy[k];
                b = id(xx, yy);
                if(xx > 0 && xx <= n && yy > 0 && yy <= m) 
                {
                    ans += x;
                    insert(a, b, x);
                    insert(b, a, x);                
                }
            }
        }
    printf("%d\n", ans - dinic());
    return 0;
}