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 1Sample Output
Case 1: 1 Case 2: 2模板题，用来测试模板，我的模板都能过。
#include<cstdio> #include<cstring> #include<queue> #define INF 1e9 using namespace std; const int maxn=200+5; struct Edge { int from,to,cap,flow; Edge() {} Edge(int f,int t,int c,int flow):from(f),to(t),cap(c),flow(flow) {} }; struct Dinic { int n,m,s,t; vector<Edge> edges; vector<int> G[maxn]; bool vis[maxn]; int cur[maxn]; int d[maxn]; void init(int n,int s,int t) { this->n=n, this->s=s, this->t=t; edges.clear(); for(int i=1; i<=n; i++) G[i].clear(); } void AddEdge(int from,int to,int cap) { edges.push_back(Edge(from,to,cap,0)); edges.push_back(Edge(to,from,0,0)); m = edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } bool BFS() { memset(vis,0,sizeof(vis)); queue<int> Q; d[s]=0; Q.push(s); vis[s]=true; while(!Q.empty()) { int x=Q.front(); Q.pop(); for(int i=0; i<G[x].size(); i++) { Edge& e=edges[G[x][i]]; if(!vis[e.to] && e.cap>e.flow) { vis[e.to]=true; Q.push(e.to); d[e.to]= 1+d[x]; } } } return vis[t]; } int DFS(int x,int a) { if(x==t || a==0) return a; int flow=0,f; for(int& i=cur[x]; i<G[x].size(); i++) { Edge& e=edges[G[x][i]]; if(d[x]+1==d[e.to] && (f=DFS(e.to,min(a,e.cap-e.flow) ))>0 ) { e.flow+=f; edges[G[x][i]^1].flow -=f; flow+=f; a-=f; if(a==0) break; } } return flow; } int Maxflow() { int flow=0; while(BFS()) { memset(cur,0,sizeof(cur)); flow += DFS(s,INF); } return flow; } } DC; int main() { int n,m,t; scanf("%d",&t); for(int i=1;i<=t;++i) { scanf("%d%d",&n,&m); DC.init(n,1,n); while(m--) { int u,v,w; scanf("%d%d%d",&u,&v,&w); DC.AddEdge(u,v,w); } printf("Case %d: %d\n",i,DC.Maxflow()); } return 0; }风骨散人Chiam 认证博客专家 拖更专业户???? 大学僧，考研狗，没上岸，ACM退役选手。名字的含义：希望可以通过努力，能力让家人拥有富足的生活而不是为了生计而到处奔波。“世人慌慌张张，不过是图碎银几两。偏偏这碎银几两，能解世间惆怅，可让父母安康，可护幼子成长 …”Chiam是 -am爱 China中国文章主要内容：Python,C++,C语言,JAVA,C#等语言的教程，ACM题解、模板、算法等，主要是数据结构，数学和图论设计模式，数据库，计算机网络，操作系统，计算机组成原理，Python爬虫、深度学习、机器学学习，计算机系408考研的所有专业课内容。目前还在更新中，博客园，微信公众号同名“风骨散人”，关注公众号可获软件大礼包
