BZOJ 1001：[BeiJing2006]狼抓兔子（最小割）

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

题意：中文。

思路：很明显是最小割，转化为最大流做。一开始看那么多点，但还是试了一下，居然过了。迷。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <iostream>
 #include <queue>
 using namespace std;
 #define N 1000010
 #define INF 0x3f3f3f3f
 struct Edge {
     int v, cap, nxt;
 } edge[N*];
 int head[N], cur[N], pre[N], tot, gap[N], dis[N], S, T;

 void Add(int u, int v, int cap) {
     edge[tot] = (Edge) {v, cap, head[u]}; head[u] = tot++;
     edge[tot] = (Edge) {u, cap, head[v]}; head[v] = tot++;
 }

 int BFS() {
     memset(gap, , sizeof(gap));
     memset(dis, INF, sizeof(dis));
     queue<int> que;
     que.push(T);
     dis[T] = ; gap[] = ;
     while(!que.empty()) {
         int u = que.front(); que.pop();
         for(int i = head[u]; ~i; i = edge[i].nxt) {
             int v = edge[i].v;
             if(INF != dis[v]) continue;
             dis[v] = dis[u] + ;
             gap[dis[v]]++;
             que.push(v);
         }
     }
 }

 int ISAP(int n) {
     BFS();
     memcpy(cur, head, sizeof(cur));
     int flow, ans = , u = pre[S] = S, i, index;
     while(dis[S] < n) {
         if(u == T) {
             flow = INF;
             for(i = S; i != T; i = edge[cur[i]].v)
                 if(flow > edge[cur[i]].cap) flow = edge[cur[i]].cap, index = i;
             for(i = S; i != T; i = edge[cur[i]].v)
                 edge[cur[i]].cap -= flow, edge[cur[i]^].cap += flow;
             ans += flow; u = index;
         }
         for(i = cur[u]; ~i; i = edge[i].nxt) if(edge[i].cap >  && dis[edge[i].v] == dis[u] - ) break;
         if(~i) {
             cur[u] = i; pre[edge[i].v] = u; u = edge[i].v;
         } else {
             if(--gap[dis[u]] == ) break;
             int md = n + ;
             for(i = head[u]; ~i; i = edge[i].nxt)
                 if(edge[i].cap >  && dis[edge[i].v] < md) md = dis[edge[i].v], cur[u] = i;
             gap[dis[u] = md + ]++;
             u = pre[u];
         }
     }
     return ans;
 }

 int main() {
     int n, m, w;
     scanf("%d%d", &n, &m);
     memset(head, -, sizeof(head));
     tot = ;
     for(int i = ; i < n; i++)
         for(int j = ; j <= m; j++)
             scanf("%d", &w), Add(j -  + i * m, j + i * m, w);
     for(int i = ; i < n; i++)
         for(int j = ; j <= m; j++)
             scanf("%d", &w), Add(j + (i - ) * m, j + i * m, w);
     for(int i = ; i < n; i++)
         for(int j = ; j <= m; j++)
             scanf("%d", &w), Add(j -  + (i - ) * m, j + i * m, w);
     S = , T = n * m;
     printf("%d\n", ISAP(T + ));
     return ;
 }