題意：K個產奶機，C頭奶牛，每個產奶機最多可供M頭奶牛使用；并告訴了產奶機、奶牛之間的兩兩距離Dij(0<=i,j<K+C)。

問題：如何安排使得在任何一頭奶牛都有自己產奶機的條件下，奶牛到產奶機的最遠距離最短？最短是多少？

解題思路：首先用Floyd把兩兩之間的最小距離算出來，接下來二分枚舉最短距離limit，只要i，j兩點之間距離小于limit，連一條容量為1的邊，源點到產奶機的邊為M,限制了每個產奶機最多供M頭奶牛。奶牛到匯點的邊為1，表示每頭牛都有一個產奶機。

#include<iostream> #include<cstdio> #include<cstring> #include<queue> using namespace std;const int maxn = 250; const int inf = 0x3f3f3f3f; struct Edge {int to,next,flow; }edge[maxn*maxn]; int n,m,c,dis[maxn][maxn]; int head[maxn],cnt; int level[maxn],st,ed;void addedge(int u,int v,int flow) {edge[cnt].to = v;edge[cnt].flow = flow;edge[cnt].next = head[u];head[u] = cnt++;swap(u,v);edge[cnt].to = v;edge[cnt].flow = 0;edge[cnt].next = head[u];head[u] = cnt++; }void floyd() {for(int k = 1; k <= n + m; k++)for(int i = 1; i <= n + m; i++)for(int j = 1; j <= n + m; j++)dis[i][j] = min(dis[i][j],dis[i][k] + dis[k][j]); }void build(int limit) {cnt = 0;memset(head,-1,sizeof(head));for(int i = 1; i <= n; i++)addedge(st,i,c);for(int i = n + 1; i <= n + m; i++)addedge(i,ed,1);for(int i = 1; i <= n; i++)for(int j = n + 1; j <= n + m; j++)if(dis[i][j] <= limit)addedge(i,j,1); }int BFS(int src,int des){queue<int> q;memset(level,0,sizeof(level));level[src] = 1;q.push(src);while(!q.empty()){int u = q.front();q.pop();if(u==des) return 1;for(int k = head[u];k!=-1;k=edge[k].next){int v = edge[k].to,w = edge[k].flow;if(level[v] == 0 && w != 0){level[v]=level[u]+1;q.push(v);}}}return -1; }int dfs(int u,int des,int increaseRoad){if(u == des) return increaseRoad;int ret = 0;for(int k=head[u];k!=-1;k=edge[k].next){int v = edge[k].to, w = edge[k].flow;if(level[v]==level[u]+1&&w!=0){int MIN = min(increaseRoad-ret,w);w = dfs(v,des,MIN);if(w > 0){edge[k].flow -=w;edge[k^1].flow +=w;ret += w;if(ret == increaseRoad) return ret;}else level[v] = -1; }}return ret; }int Dinic(int src,int des){int ans = 0;while(BFS(src,des)!=-1) ans += dfs(src,des,inf);return ans; }int main() {while(scanf("%d%d%d",&n,&m,&c)!=EOF){st = 0, ed = n + m + 1;for(int i = 1; i <= n + m; i++)for(int j = 1; j <= n + m; j++){scanf("%d",&dis[i][j]);if(dis[i][j] == 0)dis[i][j] = inf;}floyd();int l = 1,r = inf,mid,ans;while(l <= r){mid = (l + r) >> 1;build(mid);int tmp = Dinic(st,ed);if(tmp == m){ans = mid;r = mid - 1;}else l = mid + 1;}printf("%d\n",ans);}return 0; }

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