poj 1273 Drainage Ditches

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http://poj.org/problem?id=1273

最大流裸题注意测试数据有重边的处理方法即可。

EK算法代码：

#include <queue>#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const int MAX = 210;const LL INF = 1e18;LL map[MAX][MAX];queue <int> Q;int N,M,pre[MAX];bool BFS(int S, int T){///判断是否有增广路memset(pre, -1, sizeof(pre));while(!Q.empty()) Q.pop();Q.push(S);while(!Q.empty()){int u = Q.front();Q.pop();for(int i=1; i<=M; i++){if(pre[i] == -1 && map[u][i] > 0){pre[i] = u;if(i == T) return true;Q.push(i);}}}return false;}LL EK(int S, int T){///求解最大流LL maxflow = 0;while(BFS(S, T)){LL minflow = INF;for(int i=T; i!=S; i=pre[i])minflow = min(minflow, map[pre[i]][i]);for(int i=T; i!=S; i=pre[i]){map[pre[i]][i] -= minflow;map[i][pre[i]] += minflow;}maxflow += minflow;}return maxflow;}int main(){while(scanf("%d%d",&N,&M) == 2){memset(map, 0, sizeof(map));int S = 1, T = M;int u,v;LL cap;for(int i=0; i<N; i++){scanf("%d%d%lld",&u,&v,&cap);map[u][v] += cap;}printf("%lld\n",EK(S,T));}return 0;}

Dinik算法代码

#include <queue>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>using namespace std;const int MAX = 250;const int INF = 0x3f3f3f3f;int map[MAX][MAX],level[MAX],N,M;queue <int> Q;bool BFS(){///构建层次图memset(level, -1, sizeof(level));level[1] = 0;Q.push(1);while(!Q.empty()){int u = Q.front();Q.pop();for(int i=1; i<=M; i++){if(level[i] == -1 && map[u][i] > 0){level[i] = level[u] + 1;Q.push(i);}}}if(level[M] > 0) return true;return false;}int DFS(int id, int minflow){///计算增广路上最小的流量值if(id == M) return minflow;int a = 0;for(int i=1; i<=M; i++){if(map[id][i] && level[i] == level[id] + 1&& (a = DFS(i, min(map[id][i], minflow)))){map[id][i] -= a;map[i][id] += a;return a;}}return 0;}int main(){while(scanf("%d%d",&N,&M) == 2){memset(map, 0, sizeof(map));for(int i=0; i<N; i++){int u,v,cap;scanf("%d%d%d",&u,&v,&cap);map[u][v] += cap;}int tmp, ans = 0;while(BFS()){///BFS为假时代表汇点不在层次图中 无法进行增广过程while(tmp = DFS(1, INF)) ans += tmp;///累加每次增广路上增大的流量值}printf("%d\n",ans);}return 0;}

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