[2016-03-28][HDU][1074][Doing Homework]

• 时间:2016-03-28 18:46:36 星期一

• 题目编号:[2016-03-28][HDU][1074][Doing Homework]

• 题目大意:给定n门科作业的期限时间和完成耗时,每科每超过一天就扣一份,求最少扣分数

• 分析:n只有15,二进制枚举,状态压缩,枚举每种科目完成的状态,更新下一个状态,求最小值

#include <cstring>
#include <cstdio>
using namespace std;
const int maxstu = 1 << 16;
struct Cource{
        char name[20];
        int d,c;
}a[16];
struct mstatus{
        int pre,cost,reduced,id;
        mstatus(int a = 0,int b = 0,int c = 0,int d = 0):cost(a),reduced(b),pre(c),id(d){};
}dp[maxstu];
int vis[maxstu];
void print(int cur){
        if(dp[cur].pre == 0){
                printf("%s\n",a[ dp[cur].id ].name);
                return ;
        }
        print(dp[cur].pre);
        printf("%s\n",a[ dp[cur].id ].name);
}
int main(){
        int T;
        scanf("%d",&T);
        while(T--){
                int n;
                scanf("%d",&n);
                for(int i = 0;i < n;++i){
                        scanf("%s%d%d",a[i].name,&a[i].d,&a[i].c);
                }
                memset(vis,0,sizeof(vis));
                dp[0] = mstatus(0,0,0,0);
                for(int i = 0;i < (1 << n) - 1;++i){
                        for(int j = 0;j < n;++j){
                                int cur = 1 << j;
                                if((cur & i) == 0){
                                        int curcost = dp[i].cost  + a[j].c;
                                        int curreduced = dp[i].reduced + (curcost <= a[j].d?0:curcost - a[j].d);
                                        int nxtstatus = cur | i;
                                        if(vis[nxtstatus]){
                                                if(curreduced < dp[nxtstatus].reduced){
                                                        dp[nxtstatus] = mstatus(curcost,curreduced,i,j);
                                                }
                                        }else {
                                                vis[nxtstatus] = 1;
                                                dp[nxtstatus] = mstatus(curcost,curreduced,i,j);
                                        }
                                }
                        }
                }
                int ans = dp[(1 << n) - 1].reduced;
                printf("%d\n",ans);
                print((1 << n) - 1);
        }
        return 0;
}

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