Drainage Ditches
Drainage Ditches
题目链接：http://poj.org/problem?id=1273
网络流
网络流模板题，学习网络流
SAP时间复杂度为O(V*E^2)，代码如下：
1 #include <cstdio> 2 #include <iostream> 3 #include <cstring> 4 #include <vector> 5 #include <queue> 6 #define N 205 7 using namespace std; 8 typedef long long LL; 9 LL const inf=1000000000000; 10 LL mp[N][N]; 11 LL n,m; 12 bool vis[N]; 13 LL pre[N]; 14 void init(){ 15 memset(mp,0,sizeof(mp)); 16 for(LL i=0;i<n;++i){ 17 LL u,v,c; 18 cin>>u>>v>>c; 19 mp[u][v]+=c; 20 } 21 } 22 LL bfs(LL s,LL t){ 23 memset(vis,0,sizeof(vis)); 24 memset(pre,-1,sizeof(pre)); 25 queue<LL>q; 26 LL flow=inf; 27 q.push(s); 28 while(!q.empty()){ 29 LL p=q.front();q.pop(); 30 vis[p]=1; 31 if(p==t)break; 32 for(LL i=1;i<=m;++i){ 33 if(!vis[i]&&mp[p][i]!=0){ 34 flow=min(flow,mp[p][i]); 35 pre[i]=p; 36 q.push(i); 37 } 38 } 39 } 40 if(pre[t]==-1)return 0; 41 return flow; 42 } 43 void updata(LL t,LL flow){ 44 for(LL a=t;pre[a]!=-1;a=pre[a]){ 45 LL p=pre[a]; 46 mp[p][a]-=flow; 47 mp[a][p]+=flow; 48 } 49 } 50 LL getflow(LL s,LL t){ 51 LL maxflow=0,flow=0; 52 while(1){ 53 flow=bfs(s,t); 54 updata(t,flow); 55 maxflow+=flow; 56 if(flow==0)break; 57 } 58 return maxflow; 59 } 60 int main(void){ 61 while(cin>>n>>m){ 62 init(); 63 cout<<getflow(1,m)<<endl; 64 } 65 }
1 struct Edge{ 2 int from,to,cap,flow; 3 Edge(int _from=0,int _to=0,int _cap=0,int _flow=0){ 4 from=_from;to=_to;cap=_cap;flow=_flow; 5 } 6 }; 7 vector<Edge>e; 8 vector<int>adj[207]; 9 int a[207],p[207];//a[i]为可增广的权值 10 void addEdge(int from,int to,int cap){ 11 adj[from].push_back(e.size()); 12 adj[to].push_back(e.size()+1); 13 e.push_back(Edge(from,to,cap,0)); 14 e.push_back(Edge(to,from,0,0)); 15 } 16 int MaxFlow(int s,int t){ 17 int flow=0; 18 while(1){ 19 queue<int>q;//findFlow 20 q.push(s); 21 memset(a,0,sizeof(a)); 22 a[1]=INF; 23 while(!q.empty()){ 24 int u=q.front();q.pop(); 25 for(int i=0;i<adj[u].size();i++){ 26 Edge ee=e[adj[u][i]]; 27 int v=ee.to; 28 if(!a[v]&&ee.cap>ee.flow){ 29 a[v]=min(a[u],ee.cap-ee.flow); 30 p[v]=adj[u][i];//记录边的序号 31 q.push(v); 32 } 33 } 34 if(a[t])break; 35 } 36 if(!a[t])break; 37 for(int i=t;i!=1;i=e[p[i]].from){//updata残余网络 38 e[p[i]].flow+=a[t]; 39 e[p[i]^1].flow-=a[t]; 40 } 41 flow+=a[t]; 42 } 43 return flow; 44 }
