POJ 3549 Flow Problem

Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 16777    Accepted Submission(s): 7910

Problem Description

Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.

 

Input

The first line of input contains an integer T, denoting the number of test cases.
For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)
Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)

 

Output

For each test cases, you should output the maximum flow from source 1 to sink N.

 

Sample Input

23 21 2 12 3 13 31 2 12 3 11 3 1

 

Sample Output

Case 1: 1Case 2: 2
#include<stdio.h>#include<string.h>#include<algorithm>#include<queue>using namespace std;const int INF = 1e9 + 7;const int maxm = 2005;int head[maxm], V[maxm], W[maxm], Next[maxm], cur[maxm], dep[maxm];int n, m, cnt, s, t;void init(){cnt = -1, s = 1, t = n;memset(head, -1, sizeof(head));memset(Next, -1, sizeof(Next));}void add(int u, int v, int w){cnt++;Next[cnt] = head[u];head[u] = cnt;V[cnt] = v;W[cnt] = w;}int bfs(){queue<int>q;memset(dep, 0, sizeof(dep));dep[s] = 1;q.push(s);while (!q.empty()){int u = q.front();q.pop();for (int i = head[u]; i != -1;i = Next[i]){if (dep[V[i]] == 0 && W[i] != 0){dep[V[i]] = dep[u] + 1;q.push(V[i]);}} }if (dep[t] == 0) return 0;return 1;}int dfs(int u, int flow){if (u == t) return  flow;for (int i = cur[u];i != -1;i = Next[i]){if (dep[V[i]] == dep[u] + 1 && W[i] > 0){int d = dfs(V[i], min(W[i], flow));if (d > 0){W[i] -= d, W[i ^ 1] += d;return d;}}}return 0;}int dinic(){int ans = 0;while (bfs()){for (int i = 1;i <= n;i++)cur[i] = head[i];ans += dfs(s, INF);}return ans;}int main(){int  i, j, k, sum, T, x, y, z, cas = 0;scanf("%d", &T);while (T--){scanf("%d%d", &n, &m);init();for (i = 1;i <= m;i++){scanf("%d%d%d", &x, &y, &z);add(x, y, z);add(y, x, 0);}printf("Case %d: %d\n", ++cas, dinic());}return 0;}