POJ1273——最大流之Edmonds-Karp——Drainage Ditches
http://poj.org/problem?id=1273
最大流算法:
一张图有一个源点一个汇点,中间有很多要经过的点,点与点之间有最大流量的限制,问从源点到汇点的最大每秒流量是多少
加入了残余网络和增广路这两个概念
残余网络:是对于一次处理之后得到得剩下的图
增广路:对于u到i的的流量我少了x容量那么i到u我反向多个x容量,保证能够反悔不走这条路
EK算法复杂度O(VM) V:最大容量 M:边数
具体证明:http://www.cnblogs.com/luweiseu/archive/2012/07/14/2591573.html java代码看的头晕~~
代码实现:
/************************************************ * Author :Powatr * Created Time :2015-8-24 19:40:52 * File Name :POJ1273_EK.cpp ************************************************/ #include <cstdio> #include <algorithm> #include <iostream> #include <sstream> #include <cstring> #include <cmath> #include <string> #include <vector> #include <queue> #include <deque> #include <stack> #include <list> #include <map> #include <set> #include <bitset> #include <cstdlib> #include <ctime> using namespace std; #define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 typedef long long ll; const int MAXN = 2e2 + 10; const int INF = 0x3f3f3f3f; const int MOD = 1e9 + 7; int pre[MAXN]; int n, m, ans; int mp[MAXN][MAXN]; queue<int> q; void EK() { while(1){ memset(pre, 0, sizeof(pre)); while(!q.empty()) q.pop(); q.push(1); while(!q.empty()){ int now = q.front(); q.pop(); if(now == m) break; for(int i = 1; i <= m; i++){ if(mp[now][i] > 0 && !pre[i]){ pre[i] = now; q.push(i); } } } if(pre[m] == 0) break; int min1 = INF; for(int i = m; i != 1; i = pre[i]) min1 = min(min1, mp[pre[i]][i]); for(int i = m; i != 1 ; i = pre[i]){ mp[pre[i]][i] -= min1; mp[i][pre[i]] += min1; } ans += min1; } } int main(){ while(~scanf("%d%d", &n, &m)){ ans = 0; memset(mp, 0, sizeof(mp)); int u, v, val; for(int i = 1; i <= n; i++){ scanf("%d%d%d", &u, &v, &val); mp[u][v] += val; } EK(); printf("%d\n", ans); } return 0; }