題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3549

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

代碼例如以下：

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int MAXN = 32;//點數的最大值 const int MAXM = 1017;//邊數的最大值 const int INF = 0x3f3f3f3f; struct Edge {int to, cap, flow;int next; }edge[MAXM];//注意是MAXM int tol; int head[MAXN]; int dep[MAXN],pre[MAXN],cur[MAXN]; int gap[MAXN];//gap[x]=y :說明殘留網絡中dep[i]==x的個數為y void init() {tol = 0;memset(head,-1,sizeof (head)); } //加邊，單向圖三個參數。雙向圖四個參數 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++; } //輸入參數：起點、終點、點的總數 //點的編號沒有影響，僅僅要輸入點的總數 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++)//邊數{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; }

轉載于:https://www.cnblogs.com/brucemengbm/p/6846246.html

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