bzoj4145 [AMPPZ2014]The Prices

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分析：我真是弱的不行。。设f[i][j]表示前i个商店的购买状态为j的最小花销。。明显有f[i][j]=f[i-1][j]+d[i];枚举没有买的物品k，f[i][j|(1<<k-1)]=max(f[i][j]+cost[i][j]);最后还要看在i商店的买卖是否划算..f[i][j]=min(f[i][j],f[i-1][j]);

#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define fo(i,a,b) for(int i=a;i<=b;i++)#define fd(i,a,b) for(int i=a;i>=b;i--)using namespace std;int n,m,s,tot;int f[105][1<<16],c[105][105],d[105];int main(){    scanf("%d%d",&n,&m);    tot=(1<<m)-1;    fo(i,1,n)    {        scanf("%d",&d[i]);        fo(j,1,m)        {            scanf("%d",&c[i][j]);        }    }    memset(f,0x3f,sizeof(f));    f[0][0]=0;    fo(i,1,n)    {        fo(j,0,tot)f[i][j]=f[i-1][j]+d[i];        fo(k,1,m)        {            fo(j,0,tot)            if (~j&((1<<k)-1))            f[i][j|(1<<k-1)]=min(f[i][j|(1<<k-1)],f[i][j]+c[i][k]);        }        fo(j,0,tot)        f[i][j]=min(f[i][j],f[i-1][j]);    }    printf("%d\n",f[n][tot]);}

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