题目链接：
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
2 3 2 1 2 1 2 3 1 3 3 1 2 1 2 3 1 1 3 1
Sample Output
Case 1: 1 Case 2: 2
代码例如以下：
#include#include #include #include using namespace std;const int MAXN = 32;//点数的最大值const int MAXM = 1017;//边数的最大值const int INF = 0x3f3f3f3f;struct Edge{ int to, cap, flow; int next;}edge[MAXM];//注意是MAXMint tol;int head[MAXN];int dep[MAXN],pre[MAXN],cur[MAXN];int gap[MAXN];//gap[x]=y :说明残留网络中dep[i]==x的个数为yvoid init(){ tol = 0; memset(head,-1,sizeof (head));}//加边，单向图三个參数。双向图四个參数void addedge (int u,int v,int w,int rw=0){ edge[tol].to = v;edge[tol].cap = w; edge[tol].next = head[u]; edge[tol].flow = 0; head[u] = tol++; edge[tol].to = u;edge[tol].cap = rw; edge[tol]. next = head[v]; edge[tol].flow = 0;head[v]=tol++;}//输入參数：起点、终点、点的总数//点的编号没有影响，仅仅要输入点的总数int sap(int start,int end, int N){ memset(gap,0,sizeof(gap)); memset(dep,0,sizeof(dep)); memcpy(cur,head,sizeof(head)); int u = start; pre[u] = -1; gap[0] = N; int ans = 0; int i; while(dep[start] < N) { if(u == end) { int Min = INF; for( i = pre[u];i != -1; i = pre[edge[i^1]. to]) { if(Min > edge[i].cap - edge[i]. flow) Min = edge[i].cap - edge[i].flow; } for( i = pre[u];i != -1; i = pre[edge[i^1]. to]) { edge[i].flow += Min; edge[i^1].flow -= Min; } u = start; ans += Min; continue; } bool flag = false; int v; for( i = cur[u]; i != -1;i = edge[i].next) { v = edge[i]. to; if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u]) { flag = true; cur[u] = pre[v] = i; break; } } if(flag) { u = v; continue; } int Min = N; for( i = head[u]; i != -1; i = edge[i]. next) { if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min) { Min = dep[edge[i].to]; cur[u] = i; } } gap[dep[u]]--; if(!gap[dep[u]]) return ans; dep[u] = Min+1; gap[dep[u]]++; if(u != start) u = edge[pre[u]^1].to; } return ans;}int main(){ int n, m; int a, b, w; int c, s, t; int i; int T; int cas = 0; scanf("%d",&T); while(T--) { init();//初始化 scanf("%d%d",&n,&m); for(i = 1; i <= m; i++)//边数 { scanf("%d%d%d",&a,&b,&w); addedge(a,b,w,0); // addedge(b,a,w,0); } int ans = sap(1, n, n); printf("Case %d: %d\n",++cas,ans); } return 0;}