poj2516 Minimum Cost（费用流）

每种商品都跑一遍最小费用流就好了。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define N 110inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();    return x*f;}int n,m,K,h[N],num,T=100,dis[N],tot,cur[N],path[N],a[50][50],b[50][50];bool inq[N];struct edge{    int to,next,w,c;}data[6000];inline void add(int x,int y,int w,int c){    data[++num].to=y;data[num].next=h[x];h[x]=num;data[num].w=w;data[num].c=c;    data[++num].to=x;data[num].next=h[y];h[y]=num;data[num].w=0;data[num].c=-c;}inline bool spfa(){    deque<int>q;memset(dis,inf,sizeof(dis));memset(path,0,sizeof(path));    q.push_back(0);inq[0]=1;dis[0]=0;    while(!q.empty()){        int x=q.front();q.pop_front();inq[x]=0;        for(int i=h[x];i;i=data[i].next){            int y=data[i].to;if(!data[i].w) continue;            if(dis[x]+data[i].c<dis[y]){                dis[y]=dis[x]+data[i].c;path[y]=i;                if(!inq[y]){                    if(!q.empty()&&dis[y]<dis[q.front()]) q.push_front(y);                    else q.push_back(y);inq[y]=1;                }            }        }     }return path[T];}inline int solve(){    int ans=0,mxflow=0;    while(spfa()){        int low=inf,now=T;        while(path[now]) low=min(low,data[path[now]].w),now=data[path[now]^1].to;        mxflow+=low;ans+=low*dis[T];now=T;        while(path[now]) data[path[now]].w-=low,data[path[now]^1].w+=low,now=data[path[now]^1].to;    }if(mxflow!=tot) return -1;    return ans;}int main(){//  freopen("a.in","r",stdin);    while(1){        n=read();m=read();K=read();int ans=0;        if(!n&&!m&&!K) break;        for(int i=1;i<=n;++i)            for(int j=1;j<=K;++j) a[i][j]=read();        for(int i=1;i<=m;++i)            for(int j=1;j<=K;++j) b[i][j]=read();        for(int k=1;k<=K;++k){            memset(h,0,sizeof(h));num=1;tot=0;            for(int i=1;i<=n;++i)                for(int j=1;j<=m;++j) add(j+n,i,inf,read());            if(ans==-1) continue;            for(int i=1;i<=n;++i) add(i,T,a[i][k],0),tot+=a[i][k];            for(int j=1;j<=m;++j) add(0,j+n,b[j][k],0);            int res=solve();if(res==-1) ans=-1;else ans+=res;        }printf("%d\n",ans);    }return 0;}