BZOJ-1001 [BeiJing2006]狼抓兔子

解题思路：

这道题目是关于最小割的一道题目。

最小割的经典算法是根据最大流最小割定理，将最小割化成最大流然后用dinic算法求解

不过这题比较特殊，即使转换成最大流求最小割依旧不可能通过。因为时间和空间的双重限制，所以这道题的解法需要利用这个图的特殊性质。

给出的图是一个平面图无疑，那么利用平面图的特殊性质解决这个问题会简单很多。

将平面图转换成其对偶图，然后计算新的源点到新的汇点的最短路径即可得出结果。

然后关于这个题目，有一篇资料写的很好推荐一发：传送门

听说这题用dijkstra+heap速度更快。

代码：

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

typedef struct node {
	int v, cap, nxt;
	node(int a = 0, int b = 0, int c = 0) {
		v = a; cap = b; nxt = c;
	}
}Edge;

const int maxn = 2100005;
const int maxm = 6100005;
const int inf = 0x3f3f3f3f;

Edge edge[maxm];
int tot, head[maxn];
int vis[maxn], dis[maxn];

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 spfa(int s, int t) {
	int x;
	Edge e;
	queue<int> q;
	while (!q.empty()) q.pop();
	memset(vis, 0, sizeof(vis));
	memset(dis, 0x3f, sizeof(dis));
	
	dis[s] = 0; vis[s] = 1; q.push(s);
	while (!q.empty()) {
		x = q.front(); q.pop(); vis[x] = 0;
		for (int i = head[x]; ~i; i = edge[i].nxt) {
			e = edge[i];
			if (dis[x] + e.cap < dis[e.v]) {
				dis[e.v] = dis[x] + e.cap;
				if (!vis[e.v]) {
					vis[e.v] = 1;
					q.push(e.v);
				}
			}
		}
	}
	return dis[t];
}
int main() {
	int n, m, u, v, cap;
	while (~scanf("%d%d", &n, &m)) {
		if (n == 1 || m == 1) {
			if (n > m) swap(n, m);
			int ans = inf;
			for (int i = 1, a; i < m; ++i) {
				scanf("%d", &a);
				ans = min(ans, a);
			}
			printf("%d
", ans == inf ? 0 : ans);
			continue;
		}
		int s = 0, t = (m - 1) * 2 * (n - 1) + 1;
		tot = 0;
		memset(head, -1, sizeof(head));
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= m - 1; ++j) {
				scanf("%d", &cap);
				if (i == 1) { u = s; v = 2 * j; }
				else if (i == n) { u = (m - 1) * (n - 2) * 2 + 2 * j - 1; v = t; }
				else { v = (m - 1) * 2 * (i - 2) + 2 * j - 1; u = (m - 1) * 2 * (i - 1) + 2 * j; }
				add(u, v, cap);
				//printf("u = %d, v = %d, cap = %d
", u, v, cap);
			}
		}
		for (int i = 1; i < n; ++i) {
			for (int j = 1; j <= m; ++j) {
				scanf("%d", &cap);
				if (j == 1) { v = t; u = (m - 1)*(i - 1) * 2 + 1; }
				else if (j == m) { u = s; v = (m - 1) * i * 2; }
				else { u = (m - 1) * (i - 1) * 2 + 2 * (j - 1); v = u + 1; }
				add(u, v, cap);
				//printf("u = %d, v = %d, cap = %d
", u, v, cap);
			}
		}
		for (int i = 1; i < n; ++i) {
			for (int j = 1; j < m; ++j) {
				scanf("%d", &cap);
				u = (m - 1) * 2 * (i - 1) + 2 * j - 1; 
				v = u + 1;
				add(u, v, cap);
				//printf("u = %d, v = %d, cap = %d
", u, v, cap);
			}
		}
		printf("%d
", spfa(s, t));
	}
	return 0;
}