BZOJ2127 happiness

很好的网络流题目啦，只不过有点烦，不过这下总算是完全掌握了Dinic的精髓。。。

首先考虑建图：

s --> A     权值为a[A] + sigma(他和四周都选全文科的高兴值) / 2

A --> t     权值为b[A] + sigma(他和四周都选全理科的高兴值) / 2

A <--> B  权值为（同时选文科的高兴值+同时选理科的高兴值） / 2

为了解决精度问题，边权先乘以二，最后结果再除以二即可。

但是不知道为什么是对的。。。在此Orz hzwer，要是没有他的程序我就完蛋了。。。

（p.s. 作死小剧场增强版   我数组d一开始刚好开了d[10002]，然后t=10002，死活过不了，查了一下午，我也是醉了%>_<%）

/**************************************************************
    Problem: 2127
    User: rausen
    Language: C++
    Result: Accepted
    Time:2472 ms
    Memory:6900 kb
****************************************************************/
  
#include <cstdio>
#include <cstring>
#include <algorithm>
  
#define rep(i, n) for (int (i) = 1; (i) <= (n); ++(i))
#define REP for (int i = 1; i <= n; ++i) for (int j = 1; j <= m; ++j)
using namespace std;
const int inf = (int) 1e9;
  
struct edges{
    int next, to, f;
} e[500000];
  
int n, m, ans, tot = 1, s, t;
int first[10005], q[10005], d[10005];
int w[101][101], a[101][101], b[101][101];
  
inline void add_edge(int x, int y, int z){
    e[++tot].next = first[x];
    first[x] = tot;
    e[tot].to = y;
    e[tot].f = z;
}
  
inline void add_Edge(int x, int y, int z){
    add_edge(x, y, z);
    add_edge(y, x, 0);
}
  
void add_Edges(int x, int y, int z){
    add_Edge(x, y, z);
    add_Edge(y, x, z);
}
  
bool bfs(){
    memset(d, 0, sizeof(d));
    q[1] = s, d[s] = 1;
    int l = 0, r = 1, x, y;
    while (l < r){
        ++l;
        for (x = first[q[l]]; x; x = e[x].next){
            y = e[x].to;
            if (!d[y] && e[x].f)
                q[++r] = y, d[y] = d[q[l]] + 1;
        }
    }
    return d[t];
}
  
int dinic(int p, int limit){
    if (p == t || !limit) return limit;
    int x, y, tmp, rest = limit;
    for (x = first[p]; x; x = e[x].next){
        y = e[x].to;
        if (d[y] == d[p] + 1 && e[x].f && rest){
            tmp = dinic(y, min(rest, e[x].f));
            rest -= tmp;
            e[x].f -= tmp, e[x ^ 1].f += tmp;
            if (!rest) return limit;
        }
    }
    if (limit == rest) d[p] = 0;
    return limit - rest;
}
  
int Dinic(){
    int res = 0, x;
    while (bfs())
        res += dinic(s, inf);
    return res;
}
  
void make_graph(){
    int X;
    rep(i, n - 1) rep(j, m){
        scanf("%d", &X);
        ans += X, a[i][j] += X, a[i + 1][j] += X;
        add_Edges(w[i][j], w[i + 1][j], X);
    }
    rep(i, n - 1) rep(j, m){
        scanf("%d", &X);
        ans += X, b[i][j] += X, b[i + 1][j] += X;
        add_Edges(w[i][j], w[i + 1][j], X);
    }
    rep(i, n) rep(j, m - 1){
        scanf("%d", &X);
        ans += X, a[i][j] += X, a[i][j + 1] += X;
        add_Edges(w[i][j], w[i][j + 1], X);
    }
    rep(i, n) rep(j, m - 1){
        scanf("%d", &X);
        ans += X, b[i][j] += X, b[i][j + 1] += X;
        add_Edges(w[i][j], w[i][j + 1], X);
    }
    s =  n * m + 1, t = s + 1;
    REP{
        add_Edge(s, w[i][j], a[i][j]);
        add_Edge(w[i][j], t, b[i][j]);
    }
}
  
int main(){
    scanf("%d%d", &n, &m);
    REP{
        scanf("%d", a[i] + j);
        ans += a[i][j], a[i][j] <<= 1;
    }
    REP{
        scanf("%d", b[i] + j);
        ans += b[i][j], b[i][j] <<= 1;
    }
    REP w[i][j] = (i - 1) * m + j;
    make_graph();
    printf("%d
", ans - (Dinic() >> 1));
    return 0;
}