poj 2112 最大流

题意：K个产奶机，C头奶牛，每个产奶机最多可供M头奶牛使用；并告诉了产奶机、奶牛之间的两两距离Dij(0<=i,j<K+C)。

问题：如何安排使得在任何一头奶牛都有自己产奶机的条件下，奶牛到产奶机的最远距离最短？最短是多少？

二分答案。

注意先floyd求得两两之间最小距离，二分的时候上界不是200。200只是直接相连的最大距离。

  1 // File Name: 2112.cpp

  2 // Author: Missa

  3 // Created Time: 2013/4/11 星期四 14:20:52

  4 

  5 #include<iostream>

  6 #include<cstdio>

  7 #include<cstring>

  8 #include<algorithm>

  9 #include<cmath>

 10 #include<queue>

 11 #include<stack>

 12 #include<string>

 13 #include<vector>

 14 #include<cstdlib>

 15 #include<map>

 16 #include<set>

 17 using namespace std;

 18 #define CL(x,v) memset(x,v,sizeof(x));

 19 #define R(i,st,en) for(int i=st;i<en;++i)

 20 #define LL long long

 21 #define inf 0x3f3f3f3f

 22 int a[240][240];

 23 int K,C,M;

 24 

 25 const int maxn = 240;

 26 int cap[maxn][maxn];

 27 int lev[maxn];

 28 int n, m;

 29 int st, en;

 30 

 31 bool bfs()

 32 {

 33     queue <int> q;

 34     while (!q.empty()) q.pop();

 35     memset(lev, -1, sizeof(lev));

 36     lev[st] = 0;

 37     q.push(st);

 38     while (!q.empty())

 39     {

 40         int u = q.front();q.pop();

 41         for (int v = 0; v <= n+1; ++v)

 42             if (lev[v] == -1 && cap[u][v] != 0)

 43             {

 44                 lev[v] = lev[u] + 1;

 45                 q.push(v);

 46             }

 47     }

 48     return lev[en] != -1;

 49 }

 50 int dfs(int u, int cur_flow)

 51 {

 52     int dt = cur_flow;

 53     if (u == en) return dt;

 54     for (int v = 0; v <= n + 1; ++v)

 55     {

 56         if (cap[u][v] > 0 && lev[u] + 1 == lev[v])

 57         {

 58             int flow = dfs(v, min(dt, cap[u][v]));

 59             cap[u][v] -= flow;

 60             cap[v][u] += flow;

 61             dt -= flow;

 62         }

 63     }

 64     return cur_flow - dt;

 65 }

 66 

 67 int dinic()

 68 {

 69     int cur_flow, ans = 0;

 70     while(bfs())

 71         while(cur_flow = dfs(st, inf))

 72             ans += cur_flow;

 73     return ans;

 74 }

 75 void build(int dis)

 76 {

 77     memset(cap, 0, sizeof(cap));

 78     n = K + C;

 79     st = 0;

 80     en = n + 1;

 81     for (int i = 1; i <= K; ++i)

 82         cap[0][i] += M;

 83     for (int i = K + 1; i <= K + C; ++i)

 84         cap[i][en] += 1;

 85     for (int i = 1; i <= K; ++i)

 86         for (int j = K + 1; j <= K + C; ++j)

 87             if (a[i][j] <= dis)

 88                 cap[i][j] += 1;

 89 }

 90 

 91 

 92 int main()

 93 {

 94     while(~scanf("%d%d%d",&K, &C, &M))

 95     {

 96         for (int i = 1; i <= K + C; ++i)

 97             for (int j = 1; j <= K + C; ++j)

 98             {

 99                 scanf("%d", &a[i][j]);

100                 if (a[i][j] == 0) a[i][j] = inf;

101             }

102         for (int k = 1; k <= K + C; ++k)

103             for (int i = 1; i <= K + C; ++i)

104                 for (int j = 1; j <= K + C; ++j)

105                     a[i][j]= min(a[i][j],a[i][k]+a[k][j]);

106         /*for (int i = 1; i <= K + C; ++i)

107         {

108             for (int j = 1; j <= K + C; ++j)

109                 cout<<a[i][j]<<" ";

110             cout<<endl;

111         }*/

112         int ans = 1, low = 1, high = inf;

113         while(low <= high)

114         {

115             int mid = (low + high) >> 1;

116             build(mid);

117             if (dinic() == C)

118             {

119                 ans = mid;

120                 high = mid - 1;

121             }

122             else

123                 low = mid + 1;

124         }

125         printf("%d\n", ans);

126 

127     }

128     return 0;

129 }