poj 2112 floyd+Dinic最大流+二分最小值

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <queue>
  4 #define _clr(x, y) memset(x, y, sizeof(x))
  5 #define Min(x, y) (x < y ? x : y)
  6 #define INF 0x3f3f3f3f
  7 #define N 1005
  8 using namespace std;
  9 
 10 int flow[N][N], dist[N][N];
 11 int level[N];
 12 int K, C, M,  S, T, ln;
 13 
 14 void Floyd()
 15 {
 16     for(int k=1; k<=ln; k++)
 17         for(int i=1;i<=ln; i++) if(dist[i][k]<INF)
 18             for(int j=1; j<=ln; j++)
 19                 if(dist[i][k]+dist[k][j]<dist[i][j])
 20                     dist[i][j] = dist[i][k]+dist[k][j];
 21 }
 22 
 23 bool bfs()
 24 {
 25     _clr(level, -1);
 26     level[S] = 0;
 27     queue<int> Q;
 28     Q.push(S);
 29     while(!Q.empty())
 30     {
 31         int u = Q.front();
 32         Q.pop();
 33         for(int i=0; i<=T; i++)
 34         {
 35             if(flow[u][i] && level[i]<0)
 36             {
 37                 level[i] = level[u] + 1;
 38                 Q.push(i);
 39             }
 40         }
 41     }
 42     return level[T]> 0 ? 1 : 0;
 43 }
 44 
 45 int dfs(int x, int f)
 46 {
 47     int a;
 48     if(x==T) return f;
 49     for(int i=0; i<=T; i++)
 50     {
 51         if(flow[x][i] && level[i]==level[x]+1 && (a=dfs(i,Min(f,flow[x][i]))))
 52         {
 53             flow[x][i] -= a;
 54             flow[i][x] += a;
 55             return a;
 56         }
 57     }
 58     level[x] = -1;
 59     return 0;
 60 }
 61 
 62 __int64 Dinic(int len)
 63 {       
 64     //    构建残余网络
 65     _clr(flow, 0);
 66     for(int i=1; i<=K; i++)
 67         flow[S][i] = M;
 68     for(int i=K+1; i<=ln; i++)
 69         flow[i][T] = 1;
 70     for(int i=1; i<=K; i++)   // 机器
 71         for(int j=K+1; j<=ln; j++)  // 奶牛
 72             flow[i][j] = (dist[i][j]<=len);
 73 
 74     //      求解最大流
 75     __int64 ans=0, a=0;
 76     while(bfs())
 77         while(a=dfs(0,INF)) ans += a;
 78     return ans;
 79 }
 80 
 81 //  二分求解满足条件的最小解
 82 int Slove()
 83 {
 84     int l=1, r=100000;
 85     while(l<=r)
 86     {
 87         int mid = (l+r)>>1;
 88         if(Dinic(mid)>=C) r = mid-1;
 89         else l = mid+1;
 90     }
 91     return l;
 92 }
 93 int main()
 94 {
 95     while(~scanf("%d%d%d", &K, &C, &M))
 96     {
 97         ln = K+C;
 98         T = ln + 1;
 99         _clr(dist, 0);
100         for(int i=1; i<=ln; i++)
101             for(int j=1; j<=ln; j++)
102             {
103                 scanf("%d", dist[i]+j);
104                 if(dist[i][j]==0) dist[i][j]=INF;
105             }
106         //// 求出个实体之间的各个最短距离
107         Floyd();
108 
109         printf("%d\n", Slove());
110     }
111     return 0;
112 }