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[105];
    int deadline,cost;
} o[18];
struct cc
{
    int pre;
    int cost;
    int reduced;
} dp[1<<16];
bool vis[1<<16];
int tt;
void output(int now)
{
    int cur=(dp[now].pre^now);
    cur>>=1;
    int c=0;
    while(cur)
    {
        ++c;
        cur>>=1;
    }
    if(dp[now].pre!=0)
    output(dp[now].pre);
    printf("%s
",o[c].name);
}
int main()
{
    int n,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0; i<n; ++i)
            scanf("%s%d%d",o[i].name,&o[i].deadline,&o[i].cost);
        dp[0].cost=0;
        dp[0].reduced=0;
        dp[0].pre=-1;
        memset(vis,0,sizeof(vis));
        vis[0]=1;
        tt=(1<<n)-1;
        for(int i=0; i<tt; ++i)
        {
            for(int j=0; j<n; ++j)
            {

                if((i&(1<<j))==0)
                {
                    int cur=(i|(1<<j));
                    if(vis[cur])
                    {
                        int more=dp[cur].cost-o[j].deadline;
                        more=max(0,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,0);
                        vis[cur]=1;
                    }
                }
            }
        }
        printf("%d
",dp[tt].reduced);
        output(tt);
    }
    return