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就是一个求最小割.
数据比较大,n有20000,内部相连的边有20w,这么算算就要存八九十万的边,空间显然降不下来...然而打了dinic并不觉得快很多...最快跑到3800+ms
然后跪一大爷2000ms出头,他只开了50w的边这是怎么做到的qwq...然后并没有什么显著不同啊他封在一个class里(我根本不知道这玩意儿只知道跟struct差不多)...难道是读入优化打丑了...
附上他的代码地址:http://acm.hust.edu.cn/vjudge/problem/viewSource.action?id=4433085
Code:
/*========================================================================== # Last modified: 2016-03-07 19:49 # Filename: poj3469.cpp # Description: ==========================================================================*/ #define me AcrossTheSky #include <cstdio> #include <cmath> #include <ctime> #include <string> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <set> #include <map> #include <stack> #include <queue> #include <vector> #define lowbit(x) (x)&(-x) #define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++) #define FORP(i,a,b) for(int i=(a);i<=(b);i++) #define FORM(i,a,b) for(int i=(a);i>=(b);i--) #define ls(a,b) (((a)+(b)) << 1) #define rs(a,b) (((a)+(b)) >> 1) #define getlc(a) ch[(a)][0] #define getrc(a) ch[(a)][1] #define maxn 20015 #define maxm 1000500 #define pi 3.1415926535898 #define _e 2.718281828459 #define INF 1070000000 using namespace std; typedef long long ll; typedef unsigned long long ull; template<class T> inline void read(T& num) { bool start=false,neg=false; char c; num=0; while((c=getchar())!=EOF) { if(c==‘-‘) start=neg=true; else if(c>=‘0‘ && c<=‘9‘) { start=true; num=num*10+c-‘0‘; } else if(start) break; } if(neg) num=-num; } /*==================split line==================*/ int S,T,n,m; int sume=1; struct Edge{ int from,to,cap; }e[maxm]; int first[maxn],d[maxn],next[maxm],cur[maxn]; bool vis[maxn]; queue<int> q; void addedge(int x,int y,int cap){ sume++; e[sume].from=x; e[sume].to=y; e[sume].cap=cap; next[sume]=first[x]; first[x]=sume; sume++; e[sume].from=y; e[sume].to=x; e[sume].cap=0; next[sume]=first[y]; first[y]=sume; } int bfs(){ for(int i=S;i<=T;i++) vis[i]=false; q.push(0); d[0]=0; vis[0]=true; while (!q.empty()){ int now=q.front(); q.pop(); for (int i=first[now];i;i=next[i]) if (!vis[e[i].to] && e[i].cap){ d[e[i].to]=d[now]+1; vis[e[i].to]=true; q.push(e[i].to); } } return vis[T]; } int dfs(int now,int a){ if (now==T || !a) return a; int f,flow=0; for (int & i=cur[now];i;i=next[i]) if (d[now]+1==d[e[i].to] && (f=dfs(e[i].to,min(a,e[i].cap)))>0){ flow+=f; a-=f; e[i].cap-=f; e[i^1].cap+=f; if (!a) break; } return flow; } int dinic(){ int flow=0; while (bfs()){ FORP(i,0,n) cur[i]=first[i]; flow+=dfs(S,INF); } return flow; } int main(){ read(n); read(m); FORP(i,1,n) { int x; read(x); addedge(0,i,x); read(x); addedge(i,n+1,x); } FORP(i,1,m){ int x,y,z; read(x); read(y); read(z); addedge(x,y,z); addedge(y,x,z); } S=0,T=n+1; printf("%d",dinic()); }
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原文地址:http://www.cnblogs.com/YCuangWhen/p/5251971.html