/*题目描述：k个机器，每个机器最多服务m头牛。c头牛，每个牛需要1台机器来服务。告诉你牛与机器每个之间的直接距离。问：让所有的牛都被服务的情况下，使走的最远的牛的距离最短，求这个距离。分析： 首先我们可以用floyd求出各个点的最短距离d[i][j]！然后构图：建图方式s->到每头牛建边容量为1，牛到机器建边容量为1，机器到t建边容量为m如果，max_flow==c 那么 二分更新其 ans二分的关键是 大于 ans 的边 不Add进去*/#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;#define M 41000const int inf=1<<29;int gap[M],dis[M],pre[M],cur[M];int NE,NV,sink,source,d[402][402];int head[M],k,c,m,ans;struct Node{ int c,pos,next;} E[999999];#define FF(i,NV) for(int i=0;i<NV;i++)int sap(int s,int t){ //memset(pre,-1,sizeof(pre)); memset(dis,0,sizeof(int)*(NV+1)); memset(gap,0,sizeof(int)*(NV+1)); FF(i,NV) cur[i] = head[i]; int u = pre[s] = s,maxflow = 0,aug = 1<<29; gap[0] = NV; while(dis[s] < NV) {loop: for(int &i = cur[u]; i != -1; i = E[i].next) { int v = E[i].pos; if(E[i].c && dis[u] == dis[v] + 1) { aug=min(aug,E[i].c); pre[v] = u; u = v; if(v == t) { maxflow += aug; for(u = pre[u]; v != s; v = u,u = pre[u]) { E[cur[u]].c -= aug; E[cur[u]^1].c += aug; } aug = 1<<29; } goto loop; } } int mindis = NV; for(int i = head[u]; i != -1 ; i = E[i].next) { int v = E[i].pos; if(E[i].c && mindis > dis[v]) { cur[u] = i; mindis = dis[v]; } } if( (--gap[dis[u]]) == 0)break; gap[ dis[u] = mindis+1 ] ++; u = pre[u]; } return maxflow;}void addEdge(int u,int v,int c ){ E[NE].c = c; E[NE].pos = v; E[NE].next = head[u]; head[u] = NE++; E[NE].c = 0; E[NE].pos = u; E[NE].next = head[v]; head[v] = NE++;}void floyd(int n){ int i,j,k; for(k=1; k<=n; k++) { for(i=1; i<=n; i++) { if(d[i][k]!=inf) for(j=1; j<=n; j++) { if(d[k][j]!=inf&&i!=j) { d[i][j]=min(d[i][j],d[i][k]+d[k][j]); ans=max(d[i][j],ans); } } } }}int ok(int mid){ int i,j; NE=0; memset(head,-1,sizeof(int)*(NV+1)); for(i=1;i<=k;i++) addEdge(i,sink,m); for(i=k+1;i<=k+c;i++) addEdge(source,i,1); for(i=k+1;i<=k+c;i++) { for(j=1;j<=k;j++) { if(d[i][j]<=mid) addEdge(i,j,1); } } if(sap(source,sink)==c) return 1; return 0;}int main(){ int i,j; while(scanf("%d%d%d",&k,&c,&m)!=EOF) { source=ans=0; sink=k+c+1; NV=sink+1; for(i=1; i<=k+c; i++) { for(j=1; j<=k+c; j++) { scanf("%d",&d[i][j]); if(d[i][j]==0) d[i][j]=inf; } } floyd(k+c); int l=1,r=ans; while(l<=r) { int mid=(l+r)>>1; if(ok(mid)) r=mid-1; else l=mid+1; } printf("%d\n",l); } return 0;}/*2 3 20 3 2 1 13 0 3 2 02 3 0 1 01 2 1 0 21 0 0 2 0*/
