POJ 2112 Optimal Milking(最大流+二分)

题目链接

测试dinic模版，不知道这个模版到底对不对，那个题用这份dinic就是过不了。加上优化就WA，不加优化TLE。

  1 #include <cstdio>
  2 #include <string>
  3 #include <cstring>
  4 #include <queue>
  5 #include <map>
  6 #include <algorithm>
  7 using namespace std;
  8 #define INF 0x3ffffff
  9 struct node
 10 {
 11     int u,v,next,re,w;
 12 } edge[200001];
 13 int first[501],dis[501];
 14 int p[251][251];
 15 int t;
 16 int sv,ev,K,C,M;
 17 void CL()
 18 {
 19     t = 1;
 20     memset(first,-1,sizeof(first));
 21 }
 22 void add(int u,int v,int w)
 23 {
 24     edge[t].u = u;
 25     edge[t].v = v;
 26     edge[t].w = w;
 27     edge[t].re = t+1;
 28     edge[t].next = first[u];
 29     first[u] = t++;
 30     edge[t].u = v;
 31     edge[t].v = u;
 32     edge[t].w = 0;
 33     edge[t].re = t-1;
 34     edge[t].next = first[v];
 35     first[v] = t ++;
 36 }
 37 int bfs()
 38 {
 39     int u,v,i;
 40     memset(dis,0xff,sizeof(dis));
 41     queue<int> que;
 42     que.push(sv);
 43     dis[sv] = 0;
 44     while(!que.empty())
 45     {
 46         u = que.front();
 47         que.pop();
 48         for(i = first[u]; i != -1; i = edge[i].next)
 49         {
 50             v = edge[i].v;
 51             if(edge[i].w > 0&&dis[v] < 0)
 52             {
 53                 dis[v] = dis[u] + 1;
 54                 que.push(v);
 55             }
 56         }
 57     }
 58     if(dis[ev] > 0) return 1;
 59     else return 0;
 60 }
 61 int dfs(int u,int step)
 62 {
 63     int i,a = 0,v,flag = 0;
 64     if (u == ev) return step;
 65     for (i = first[u];i != -1&&flag < step; i = edge[i].next)//flag<step
 66     {
 67         v = edge[i].v;
 68         if (edge[i].w > 0&& dis[v] == dis[u]+1&&(a = dfs(v,min(step,edge[i].w))))
 69         {
 70             edge[i].w -= a;
 71             flag += a;//这里
 72             edge[edge[i].re].w += a;
 73             return a;
 74         }
 75     }
 76     if(flag == 0) dis[u] = -1;//这里
 77     return flag;
 78 }
 79 void build(int x)
 80 {
 81     int i,j;
 82     CL();
 83     for(i = 1; i <= K; i ++)
 84     {
 85         add(0,i,M);
 86     }
 87     for(i = 1; i <= K; i ++)
 88     {
 89         for(j = 1; j <= C; j ++)
 90         {
 91             if(p[i][j+K] <= x)
 92             add(i,K+j,1);
 93         }
 94     }
 95     for(i = 1;i <= C;i ++)
 96     {
 97         add(i+K,ev,1);
 98     }
 99 }
100 int fun(int x)
101 {
102     int ans = 0,res;
103     build(x);
104     while(bfs())
105     {
106           while(res=dfs(sv,INF))
107           ans += res;
108     }
109     if(ans == C)
110         return 1;
111     else
112         return 0;
113 }
114 int bin(int l,int r)
115 {
116     int str,mid,end;
117     str = l;
118     end = r;
119     while(str < end)
120     {
121         mid = (str + end)/2;
122         if(fun(mid))
123         {
124             end = mid;
125         }
126         else
127         {
128            str = mid + 1;
129         }
130     }
131     return end;
132 }
133 int main()
134 {
135     int i,j,k;
136     while(scanf("%d%d%d",&K,&C,&M)!=EOF)
137     {
138         for(i = 1;i <= K+C;i ++)
139         {
140             for(j = 1;j <= K+C;j ++)
141             {
142                 scanf("%d",&p[i][j]);
143                 if(i != j&&p[i][j] == 0)
144                 p[i][j] = INF;
145             }
146         }
147         for(i = 1;i <= K+C;i ++)
148         {
149             for(j = 1;j <= K+C;j ++)
150             {
151                 for(k = 1;k <= K+C;k ++)
152                 {
153                     if(p[j][k] > p[j][i] + p[i][k])
154                     p[j][k] = p[j][i] + p[i][k];
155                 }
156             }
157         }
158         sv = 0;
159         ev = K+C+1;
160         int maxz = 0;
161         for(i = 1;i <= K+C;i ++)
162         {
163             for(j = 1;j <= K+C;j ++)
164             maxz = max(maxz,p[i][j]);
165         }
166         printf("%d\n",bin(0,maxz));
167     }
168     return 0;
169 }

 

 

posted @ 2013-08-08 09:40  Naix_x  阅读(186)  评论(0)  编辑  收藏  举报