标签:oid char output define 题目 pop c++ read dfs
题目大意:略。
解题思路:最大流木板题,以下是Dinic算法的代码。
C++ Code:
#include<stdio.h> #include<cctype> #include<vector> #include<algorithm> #include<cstring> #include<queue> using namespace std; #define INF 0x3f3f3f3f char buf[40000020]; int bufpos,n,m,s,t,level[10002],iter[10002]; struct edges{ int to,cap,rev; edges(int t,int c,int r):to(t),cap(c),rev(r){} }; vector<edges>G[10002]; queue<int>q; inline void init(){ buf[fread(buf,1,40000000,stdin)]=‘\0‘; bufpos=0; } inline int readint(){ int p=0; for(;!isdigit(buf[bufpos]);bufpos++); for(;isdigit(buf[bufpos]);bufpos++) p=p*10+buf[bufpos]-‘0‘; return p; } inline void addedge(int from,int to,int cap){ G[from].push_back(edges(to,cap,G[to].size())); G[to].push_back(edges(from,0,G[from].size()-1)); } void bfs(int s){ memset(level,-1,sizeof(level)); level[s]=0; q.push(s); while(!q.empty()){ int u=q.front(); q.pop(); for(int i=0;i<G[u].size();++i){ edges& e=G[u][i]; if(level[e.to]<0&&e.cap>0){ level[e.to]=level[u]+1; q.push(e.to); } } } } int dfs(int u,int t,int f){ if(u==t)return f; for(int& i=iter[u];i<G[u].size();++i){ edges& e=G[u][i]; if(e.cap>0&&level[e.to]>level[u]){ int d=dfs(e.to,t,min(f,e.cap)); if(d){ e.cap-=d; G[e.to][e.rev].cap+=d; return d; } } } return 0; } int max_flow(int s,int t){ int flow=0; while(1){ bfs(s); if(level[t]<0)return flow; memset(iter,0,sizeof(iter)); int f; while(f=dfs(s,t,INF))flow+=f; } } int main(){ #ifdef DEBUG freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif init(); n=readint(),m=readint(),s=readint(),t=readint(); for(int i=1;i<=m;++i){ int u=readint(),v=readint(),cap=readint(); addedge(u,v,cap); } printf("%d\n",max_flow(s,t)); return 0; }
标签:oid char output define 题目 pop c++ read dfs
原文地址:http://www.cnblogs.com/Mrsrz/p/7254062.html