poj2516（最小費用最大流模版）

首先就這題而言，各k互不影響，可以分開來算，累加即可

對於構圖來說，以後可以源點s=0，匯點t=n+m，其中一組點是1~n，另一組n+1~n+m，注意可以這樣構圖

對於最小費用最大流，要用spfa來做最短路徑，只需記錄路徑即可

#include#include#include#include#include#define n 130

using namespace std;

const int inf=0x3f3f3f3f;

int n,m,k,offer[n][n],need[n][n],map[n][n],cost[n][n],sumn[n],sumo[n],pre[n],s,t;

bool spfa()

; int dis[n];

memset(dis,inf,sizeof(dis));//對了，不能int dis=｛inf｝這樣付初值，這樣絕對不可以！！！！！！！！！！！！！

queueq;

vis[s]=true;

dis[s]=0;

q.push(s);

while (!q.empty())

}vis[u]=false;

}/spfa，

if (dis[t]==inf) return false;else return true;//判斷是否有增廣路，然後返回即可。

}int work()

return ans;//返回最小費用！

}int main()

{ while (scanf("%d%d%d",&n,&m,&k))

{ memset(sumn,0,sizeof(sumn));

memset(sumo,0,sizeof(sumo));

memset(need,0,sizeof(need));

memset(offer,0,sizeof(offer));

if (n==0) break;

int sign=0;

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

for (int j=1;j<=k;j++) scanf("%d",&need[i][j]),sumn[j]+=need[i][j];

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

for (int j=1;j<=k;j++) scanf("%d",&offer[i][j]),sumo[j]+=offer[i][j];

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

if (sumo[j]

最小費用最大流poj2516

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poj2516 最小費用最大流

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