Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch. Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network. Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.
The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.
For each case, output a single integer, the maximum rate at which water may emptied from the pond.
模板题，题意很明了，直接测板子。
#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; while(scanf("%d%d",&m,&n)==2){ DC.init(n,1,n); while(m--) { int u,v,w; scanf("%d%d%d",&u,&v,&w); DC.AddEdge(u,v,w); } printf("%d\n",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
