poj 2455(二分+最大流)
题意:从1到n,要走不少于K条路,每条路不能重复走,点可以重复走,求所走的路中最长的最小,,
找出距离的最大值,然后二分,建图时符合的每条边链接的两点间建双向边,流量都为1,求最大流
#include<stdio.h> #include<string.h> const int N=210; const int inf=0x3fffffff; int dis[N],gap[N],head[N],num,start,end,ans,n,m; struct edge { int st,ed,flow,next; }E[200000]; struct node { int x,y,w; }P[40000]; void addedge(int x,int y,int w) { E[num].st=x;E[num].ed=y;E[num].flow=w;E[num].next=head[x];head[x]=num++; E[num].st=y;E[num].ed=x;E[num].flow=w;E[num].next=head[y];head[y]=num++; } void makemap(int D) { memset(head,-1,sizeof(head)); num=0; int i; for(i=0;i<m;i++) { if(P[i].w<=D) addedge(P[i].x,P[i].y,1); } } int dfs(int u,int minflow) { if(u==end)return minflow; int i,v,f,flow=0,min_dis=ans-1; for(i=head[u];i!=-1;i=E[i].next) { if(E[i].flow) { v=E[i].ed; if(dis[v]+1==dis[u]) { f=dfs(v,E[i].flow>minflow-flow?minflow-flow:E[i].flow); E[i].flow-=f; E[i^1].flow+=f; flow+=f; if(flow==minflow)break; if(dis[start]>=ans)return flow; } min_dis=min_dis>dis[v]?dis[v]:min_dis; } } if(flow==0) { if(--gap[dis[u]]==0) dis[start]=ans; dis[u]=min_dis+1; gap[dis[u]]++; } return flow; } int isap() { int maxflow=0; memset(dis,0,sizeof(dis)); memset(gap,0,sizeof(gap)); gap[0]=ans; while(dis[start]<ans) maxflow+=dfs(start,inf); return maxflow; } int main() { int i,k,L,R,mid; while(scanf("%d%d%d",&n,&m,&k)!=-1) { R=0;start=1;end=n;ans=n; for(i=0;i<m;i++) { scanf("%d%d%d",&P[i].x,&P[i].y,&P[i].w); if(R<P[i].w)R=P[i].w; } L=0; while(L<R) { mid=(L+R)/2; makemap(mid); if(isap()>=k) R=mid; else L=mid+1; } printf("%d\n",R); } return 0; }
补充:软件开发 , C++ ,