POJ1273 Drainage Ditches 最大流裸题

　　最大流的裸题，纯水，EK都可以秒过。今天敲了一下ISAP，发现还不是很方便，感觉Dinic好敲多了，而且Dinic的效率又比较稳定，以后如果不是碰到那种特大的数据，还是敲Dinic吧。

  1 //STATUS:C++_AC_0MS_760KB
  2 #include<stdio.h>
  3 #include<stdlib.h>
  4 #include<string.h>
  5 #include<math.h>
  6 #include<iostream>
  7 #include<string>
  8 #include<algorithm>
  9 #include<vector>
 10 #include<queue>
 11 #include<stack>
 12 #include<map>
 13 using namespace std;
 14 #define LL long long
 15 #define Max(a,b) ((a)>(b)?(a):(b))
 16 #define Min(a,b) ((a)<(b)?(a):(b))
 17 #define mem(a,b) memset(a,b,sizeof(a))
 18 #define lson l,mid,rt<<1
 19 #define rson mid+1,r,rt<<1|1
 20 const int MAX=210,INF=0x3f3f3f3f;
 21 
 22 struct Edge{
 23     int from,to,cap,flow;
 24 };
 25 
 26 vector<Edge> edge;
 27 vector<int> g[MAX];
 28 int d[MAX],cur[MAX],vis[MAX];
 29 int n,m,s,t;
 30 
 31 int bfs()
 32 {
 33     int x,i;
 34     queue<int> q;
 35     q.push(s);
 36     mem(vis,0);
 37     d[s]=0;
 38     vis[s]=1;
 39     while(!q.empty())
 40     {
 41         x=q.front();q.pop();
 42         for(i=0;i<g[x].size();i++){
 43             Edge& e=edge[g[x][i]];
 44             if(!vis[e.to] && e.cap>e.flow){
 45                 vis[e.to]=1;
 46                 d[e.to]=d[x]+1;
 47                 q.push(e.to);
 48             }
 49         }
 50     }
 51     return vis[t];
 52 }
 53 
 54 int dfs(int x,int a)
 55 {
 56     if(x==t || a==0)return a;
 57     int f,flow=0;
 58     for(int& i=cur[x];i<g[x].size();i++){
 59         Edge& e=edge[g[x][i]];
 60         if(d[x]+1==d[e.to] && (f=dfs(e.to,Min(a,e.cap-e.flow)))>0){
 61             e.flow+=f;
 62             edge[g[x][i]^1].flow-=f;
 63             flow+=f;
 64             a-=f;
 65             if(!a)break;
 66         }
 67     }
 68     return flow;
 69 }
 70 
 71 int dinic()
 72 {
 73     int flow=0;
 74     while(bfs()){
 75         mem(cur,0);
 76         flow+=dfs(s,INF);
 77     }
 78     return flow;
 79 }
 80 
 81 int main()
 82 {
 83  //   freopen("in.txt","r",stdin);
 84     int i,j,u,v,val,ans;
 85     while(~scanf("%d%d",&m,&n))
 86     {
 87         s=1,t=n;
 88         for(i=1;i<=n;i++)g[i].clear();
 89         edge.clear();
 90         m<<=1;
 91         for(i=0;i<m;i+=2){
 92             scanf("%d%d%d",&u,&v,&val);
 93             edge.push_back((Edge){u,v,val,0});
 94             edge.push_back((Edge){v,u,0,0});
 95             g[u].push_back(i);
 96             g[v].push_back(i+1);
 97         }
 98 
 99         ans=dinic();
100 
101         printf("%d\n",ans);
102     }
103     return 0;
104 }