HDU 1074 Doing Homework 状压DP

由于数据量较小，直接二进制模拟全排列过程，进行DP，思路由kuangbin blog得到，膜拜斌神

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
struct asd
{
    char name[];
    int deadline,cost;
} o[];
struct cc
{
    int pre;
    int cost;
    int reduced;
} dp[<<];
bool vis[<<];
int tt;
void output(int now)
{
    int cur=(dp[now].pre^now);
    cur>>=;
    int c=;
    while(cur)
    {
        ++c;
        cur>>=;
    }
    if(dp[now].pre!=)
    output(dp[now].pre);
    printf("%s\n",o[c].name);
}
int main()
{
    int n,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=; i<n; ++i)
            scanf("%s%d%d",o[i].name,&o[i].deadline,&o[i].cost);
        dp[].cost=;
        dp[].reduced=;
        dp[].pre=-;
        memset(vis,,sizeof(vis));
        vis[]=;
        tt=(<<n)-;
        for(int i=; i<tt; ++i)
        {
            for(int j=; j<n; ++j)
            {

                if((i&(<<j))==)
                {
                    int cur=(i|(<<j));
                    if(vis[cur])
                    {
                        int more=dp[cur].cost-o[j].deadline;
                        more=max(,more)+dp[i].reduced;
                        if(more<dp[cur].reduced)
                            dp[cur].pre=i,dp[cur].reduced=more;

                    }
                    else
                    {
                        dp[cur].pre=i;
                        dp[cur].cost=dp[i].cost+o[j].cost;
                        int more=dp[cur].cost-o[j].deadline;
                        dp[cur].reduced=dp[i].reduced+max(more,);
                        vis[cur]=;
                    }
                }
            }
        }
        printf("%d\n",dp[tt].reduced);
        output(tt);
    }
    return