标签:txt inline turn bool 最短路 log open online cstring
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1001
顺便推荐一个ppt,里面有对平面图的介绍:浅析最大最小定理在信息学竞赛中的应用。
这里直接求最小割肯定会T,所以应把原图看成一张平面图,ppt中说该平面图对应的对偶图的每一个环对应原图的一个割,这点有些不理解,不过不影响做这一道题。想象一下,在最外面那个无限大的平面,由左上角朝右下角连一条附加的边,这么做就多了一个附加面,设这条附加的边的权值为 -inf,那么最小割一定包含这一条边。把这条边去掉,就成了求一个最短路的问题了。
#include <cstdio> #include <cstring> const int maxn = 1005, maxnd = maxn * maxn << 1, maxe = maxn * maxn * 3; int n, m, S, T, special = 2147483647, t1, t2, t3; int head[maxnd], to[maxe << 1], next[maxe << 1], w[maxe << 1], lb; char ch; bool inq[maxnd]; int que[maxnd], h, head_, tail, d[maxnd]; inline void ist(int aa, int ss, int ww) { to[lb] = ss; next[lb] = head[aa]; head[aa] = lb; w[lb] = ww; ++lb; } inline void readint(int & rt) { while ((ch = getchar()) < 48); rt = ch - 48; while ((ch = getchar()) > 47) { rt = rt * 10 + ch - 48; } } int main(void) { //freopen("in.txt", "r", stdin); memset(head, -1, sizeof head); memset(next, -1, sizeof next); readint(n); readint(m); if (n == 1 || m == 1) { while (scanf("%d", &t1) != EOF) { special = special < t1? special: t1; } printf("%d\n", special); return 0; } S = (n - 1) * (m - 1) * 2 + 1; T = S + 1; for (int i = 1; i <= n; ++i) { for (int j = 1; j < m; ++j) { t2 = (i - 1) * (m - 1) * 2 + j * 2; t1 = t2 - (m - 1) * 2 - 1; t1 = t1 > 0? t1: T; t2 = t2 < S? t2: S; scanf("%d", &t3); ist(t1, t2, t3); ist(t2, t1, t3); } } for (int i = 1; i < n; ++i) { t2 = (i - 1) * (m - 1) * 2 + 1; scanf("%d", &t3); ist(S, t2, t3); ist(t2, S, t3); for (int j = 2; j < m; ++j) { t1 = (i - 1) * (m - 1) * 2 + (j - 1) * 2; t2 = t1 + 1; scanf("%d", &t3); ist(t1, t2, t3); ist(t2, t1, t3); } t1 = i * (m - 1) * 2; scanf("%d", &t3); ist(t1, T, t3); ist(T, t1, t3); } for (int i = 1; i < n; ++i) { for (int j = 1; j < m; ++j) { t2 = (i - 1) * (m - 1) * 2 + j * 2; t1 = t2 - 1; scanf("%d", &t3); ist(t1, t2, t3); ist(t2, t1, t3); } } memset(d, 0x3c, sizeof d); que[tail++] = S; inq[S] = true; d[S] = 0; while (head_ != tail) { h = que[head_++]; if (head_ == T) { head_ = 0; } inq[h] = false; for (int j = head[h]; j != -1; j = next[j]) { if (d[to[j]] > d[h] + w[j]) { d[to[j]] = d[h] + w[j]; if (!inq[to[j]]) { inq[to[j]] = true; que[tail++] = to[j]; if (tail == T) { tail = 0; } } } } } printf("%d\n", d[T]); return 0; }
_bzoj1001 [BeiJing2006]狼抓兔子【平面图】
标签:txt inline turn bool 最短路 log open online cstring
原文地址:http://www.cnblogs.com/ciao-sora/p/6146211.html