poj2516Minimum Cost

http://poj.org/problem?id=2516

建图的时候 有个地方写错了 卡了半年。。

题意看了N久啊 有N个店主需要K种物品 有M个供应点 每个供应点有K种物品 其实是算K次最小费用 然后叠加 分解开来这题就是求把某种物品从供应点送到店主那里

多个源点-》多个汇点  所以加一个超级源点 和 超级汇点 源点->M个供应点->N个店主->汇点 所以共有m+n+2条边 套最小费用模板就行了

View Code

  1 #include <iostream>

  2 #include<cstring>

  3 #include<cstdio>

  4 #include<algorithm>

  5 #include<queue>

  6 #define MN 10000

  7 #define MM 10000

  8 #define INF 0xfffffff

  9 using namespace std;

 10 struct node

 11 {

 12     int u,v,cost,cap,next;

 13 }edge[MM];

 14 int head[MN],t,vis[MN],dist[MN],co[55][55][55],sh[55][55],sto[55][55],pp[MN],flow;

 15 void init()

 16 {

 17     t = 0;

 18     memset(head,-1,sizeof(head));

 19 }

 20 void add(int u,int v,int c,int p)

 21 {

 22     edge[t].u = u;

 23     edge[t].v = v;

 24     edge[t].cap = c;

 25     edge[t].cost = p;

 26     edge[t].next = head[u];

 27     head[u] = t++;

 28     edge[t].v = u;

 29     edge[t].u = v;

 30     edge[t].cap = 0;

 31     edge[t].cost = -p;

 32     edge[t].next = head[v];

 33     head[v] = t++;

 34 }

 35 int spfa(int s,int en,int n)

 36 {

 37     int i,j,u,v;

 38     for(i =0 ; i <= n ; i++)

 39     dist[i] = INF;

 40     memset(vis,0,sizeof(vis));

 41     memset(pp,-1,sizeof(pp));

 42     queue<int>q;

 43     q.push(s);

 44     vis[s] = 1;

 45     dist[s] = 0;

 46     while(!q.empty())

 47     {

 48         u = q.front();

 49         q.pop();

 50         vis[u] = 0;

 51         for(i = head[u] ; i != -1 ; i = edge[i].next)

 52         {

 53             v = edge[i].v;

 54             if(edge[i].cap&&dist[v]>dist[u]+edge[i].cost)

 55             {

 56                 dist[v] = dist[u]+edge[i].cost;

 57                 //cout<<dist[v]<<endl;

 58                 pp[v] = i;

 59                 if(!vis[v])

 60                 {

 61                     vis[v] = 1;

 62                     q.push(v);

 63                 }

 64             }

 65         }

 66     }

 67     //cout<<dist[en]<<" ";

 68     if(dist[en]==INF)

 69     return 0;

 70     return 1;

 71 }

 72 int mcmf(int s,int en,int n)

 73 {

 74     int i,minf,minc=0;

 75     while(spfa(s,en,n))

 76     {

 77         minf = INF+1;

 78         for(i = pp[en] ; i != -1 ; i = pp[edge[i].u])

 79         {

 80             if(edge[i].cap<minf)

 81             minf = edge[i].cap;

 82         }

 83         flow+=minf;

 84         for(i = pp[en] ; i != -1 ; i = pp[edge[i].u])

 85         {

 86             edge[i].cap-=minf;

 87             edge[i^1].cap+=minf;

 88         }

 89         minc+=minf*dist[en];

 90     }

 91     return minc;

 92 }

 93 int main()

 94 {

 95     int i,j,n,m,k,g;

 96     while(cin>>n>>m>>k)

 97     {

 98         if(n==0&&m==0&&k==0)

 99         break;

100         int flag = 0,ss=0;

101         for(i = 1; i <= n ; i++)

102             for(j = 1; j <= k ; j++)

103                 cin>>sh[i][j];

104         for(i = 1; i <= m ; i++)

105              for(j = 1; j <= k ; j++)

106                 cin>>sto[i][j];

107         for(i = 1;i <= k ; i++)

108             for(j = 1 ; j <= n ; j++)

109                 for(g = 1; g <= m ; g++)

110                 cin>>co[i][j][g];

111         int s = 0,en = n+m+1;

112         for(i = 1; i <= k ; i++)

113         {

114             init();

115             int sa=0;

116             for(j = 1 ; j <= n ; j++)

117             {

118                add(m+j,en,sh[j][i],0);

119                sa+=sh[j][i];

120             }

121             for(j = 1; j <= m ; j++)

122                 add(s,j,sto[j][i],0);

123             for(j = 1; j <= m ; j++)

124                 for(g = 1 ; g <= n ; g++)

125                     add(j,m+g,INF,co[i][g][j]);

126             flow=0;

127             ss+=mcmf(s,en,n+m+1);

128             if(sa>flow)

129                 break;

130         }

131         if(i==k+1)

132         cout<<ss<<endl;

133         else

134         cout<<"-1\n";

135     }

136     return 0;

137 }