HDU 1074 Doing Homework

题意： 输入t表示样例数目，输入n表示有n个科目，输入n行，每行有 name，endtime，costtime。超过endtime的一个扣一分，求需要扣最少的分数。

分析：压缩状态，每次状态都可能由很多种状态转化过来，找出扣分最少的，更新当前状态的值。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>

using namespace std;

struct subject
{
    char Name[111];
    int End, Cost;
} S[1<<16];
struct DP
{
    int cost;
    int reduced;
    int pre;
} dp[1<<16];
int visit[1<<16];

void print(int date)
{
    if(date == 0)
    {
        return ;
    }
    int which;
    which = date^dp[date].pre;
    int number=0;
    while(which)
    {
        which>>=1;
        number++;
    }
    print(dp[date].pre);
    printf("%s\n", S[number-1].Name);
}

int main()
{
    int t, n;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &n);
        for(int i=0; i<n; i++)
        {
            scanf(" %s %d %d", S[i].Name, &S[i].End, &S[i].Cost);
        }
        int e = (1<<n)-1;
        dp[0].cost=0;
        dp[0].pre=-1;
        dp[0].reduced = 0;
        visit[0] = 1;
        memset(visit, 0, sizeof(visit));
        for(int i=0; i<e; i++)
        {
            for(int j=0; j<n; j++)
            {
                int date1 = 1<<j;
                if( (i&date1)==0)//这个作业没做过
                {
                    int date2 = i|date1;
                    dp[date2].cost = dp[i].cost + S[j].Cost;
                    int reduce = dp[date2].cost - S[j].End;
                    if(reduce<0)
                        reduce = 0;
                    reduce += dp[i].reduced;

                    if(visit[date2])//之前状态存在过
                    {
                        if(reduce<dp[date2].reduced)
                        {
                            dp[date2].reduced = reduce;
                            dp[date2].pre = i;
                        }
                    }
                    else
                    {
                        visit[date2] = 1;
                        dp[date2].reduced = reduce;
                        dp[date2].pre = i;
                    }

                }
            }
        }
        printf("%d\n", dp[e].reduced);
        print(e);
    }

    return 0;
}