POJ 2391
来源:互联网 发布:淘宝批量编辑宝贝 编辑:程序博客网 时间:2024/05/15 16:17
求最大值最小问题通常都是用二分.
#include<iostream>#include<cstdio>#include<cstring>#define min(a,b) ((a)>(b))?(b):(a)#define inf 1000100000#define INF (~(0ULL)) >>1using namespace std;int N,n,m;long long map[205][205];int in[205],out[205];int k,dist[510],gap[510],edgeHead[500];//双向边struct{ int v, cap, next, re;}edge[500500];void addEdge(int u,int v,int ca){ edge[k].v=v; edge[k].cap=ca; edge[k].next=edgeHead[u]; edge[k].re=k + 1; //这个用来记录此边的反边 edgeHead[u]=k ++; edge[k].v=u; edge[k].cap=0; edge[k].next=edgeHead[v]; edge[k].re=k - 1; edgeHead[v]=k ++;}void build_graph(long long mid){int i,j;memset(edgeHead,0,sizeof(edgeHead));k=1;for(i=1;i<=n;i++){addEdge(0,i,in[i]);addEdge(i+n,2*n+1,out[i]);for(j=1;j<=n;j++)if(map[i][j]<=mid)addEdge(i,j+n,inf);}N=2*n+1;}int dfs (int p , int limit){ if(p==N)return limit; for(int i=edgeHead[p];i!=0;i=edge[i].next){ int v = edge[i].v; if(dist[p]==(dist[v]+1) && edge[i].cap>0){ int t=dfs(v,min(limit , edge[i].cap)); if(t<0)return t;//没有增广路了 if(t>0)//更新流 { edge[i].cap-=t; edge[edge[i].re].cap+=t; return t; } } } int tmp=N+1; for(int i=edgeHead[p];i!=0;i=edge[i].next){ int v = edge[i].v; if(edge[i].cap>0) tmp=min(tmp,dist[v]+1); } if(--gap[dist[p]]==0 || dist[0]>N)return -1;//出现断层或回流已满 ++gap[dist[p]=tmp]; return 0;}int SAP(){ gap[0]=N+1; int f = 0 , t=0; while (~(t=dfs(0,inf))) f+=t; return f;}int main(){int i,j,sum=0;scanf("%d %d",&n,&m);for(i=1;i<=n;i++){scanf("%d %d",&in[i],&out[i]);sum+=in[i];for(j=1;j<=n;j++)map[i][j]=INF;map[i][i]=0;}for(i=1;i<=m;i++){int u,v,w;scanf("%d %d %d",&u,&v,&w);map[u][v]=map[u][v]>w?w:map[u][v];map[v][u]=map[v][u]>w?w:map[v][u];}long long l=0,h=0;for(k=1;k<=n;k++)for(i=1;i<=n;i++)if(map[i][k]!=INF)for(j=1;j<=n;j++)if(map[k][j]!=INF){if(map[i][j]>map[i][k]+map[k][j]){map[i][j]=map[i][k]+map[k][j];}}for(i=1;i<=n;i++)for(j=1;j<=n;j++)if(map[i][j]!=INF){h=h>map[i][j]?h:map[i][j];}long long ans=-1;while(l<=h){long long mid=(l+h)>>1;build_graph(mid);memset(dist,0,sizeof(dist));memset (gap , 0 , sizeof(gap));int kk=SAP();if(kk==sum){h=mid-1;ans=mid;}elsel=mid+1;}printf("%lld\n",ans);return 0;}
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