【最大流】北大 poj 1273 Drainage Ditches

/* THE PROGRAM IS MADE BY PYY *//*----------------------------------------------------------------------------//Copyright (c) 2011 panyanyany All rights reserved.URL   : http://poj.org/problem?id=1273Name  : 1273 Drainage DitchesDate  : Sunday, January 1, 2012Time Stage : seven to eight hours aroundResult: 9696192panyanyany1273Accepted348K16MSC++1974B2012-01-01 15:55:419696127panyanyany1273Wrong AnswerC++2316B2012-01-01 15:03:349696116panyanyany1273Wrong AnswerC++2319B2012-01-01 14:55:28Test Data :Review :最大流，看了许久，大部分都可以慢慢看懂，唯一难搞的就是增加反向边的问题，看到大牛的例子，才明白反向边确实很重要。一开始看的是这位大牛的文章：http://www.cppblog.com/mythit/archive/2009/04/19/80470.aspx这篇文章则给出了例子，证明反向边确实重要：http://pictureyong.blog.163.com/blog/static/851870682011299044852/这位大牛的解释挺详细的：http://blog.csdn.net/akof1314/article/details/4845183听华神说这种方法是最慢的，当时有脑没脑地想了一下，跟深搜比起来，广搜应该是很快的了。后来突然意识到，这种算法不止搜了一次！也就是说，如果一张图很大的话，这种算法可能会需要对其反复搜索，而每次大范围的搜索却又只找到一条路，如果路不对，还要继续找……于是这样就会很费时间了……呃，好吧，我也是刚学，进阶的事慢慢来……//----------------------------------------------------------------------------*/#include <stdio.h>#include <string.h>#include <queue>using namespace std ;#define DEBUG#define min(x, y)((x) < (y) ? (x) : (y))#define INF0x3f3f3f3f#define MAXN202intn, m ;intmap[MAXN][MAXN], path[MAXN], flow[MAXN] ;int Mark_Point (const int start, const int end){intt, i, j ;queue<int>q ;memset (path, 0, sizeof (path)) ;flow[start] = INF ;q.push (start) ;//puts ("---------------------Loop:") ;while (!q.empty ()){t = q.front () ;q.pop () ;if (end == t)break ;//printf ("Point: %d\n", t) ;for (i = 1 ; i <= end ; ++i){if ((i != start) && map[t][i] && !path[i]){path[i] = t ;flow[i] = min (flow[t], map[t][i]) ;//printf ("\tPoint: %d, (%d, %d)\n", i, path[i], flow[i]) ;q.push (i) ;}}}if (path[end] == 0)return -1 ;return flow[end] ;}int Edmonds_Karp (const int start, const int end){int incr, step, curr, prev ;incr = 0 ;while ((step = Mark_Point (start, end)) != -1){incr += step ;curr = end ;while (curr != start){prev = path[curr] ;map[prev][curr] -= step ;map[curr][prev] += step ;//printf ("Point: %d --> Point: %d === %d\n", prev, curr, //map[prev][curr]) ;curr = prev ;}}return incr ;}int main (void){inti, j ;int x, y, c ;while (~scanf ("%d%d", &n, &m)){memset (map, 0, sizeof (map)) ;for (i = 0 ; i < n ; ++i){scanf ("%d%d%d", &x, &y, &c) ;map[x][y] += c ;}printf ("%d\n", Edmonds_Karp (1, m)) ;}return 0 ;}