POJ-1273 Drainage Ditches 最大流裸题
题目链接:http://poj.org/problem?id=1273
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 }