HDU 1074 Doing Homework 狀壓dp

狀態數不多，做到每一個作業所用的罰時和前面已經做了哪些作業密切相關，還有其他一些性質都和狀壓dp非常契合，那麼就是他了。

#include<iostream>
#include<cstdio>
#include<string.h>
#include<algorithm>
#include<map>
#include<string>
using namespace std;

int t,n,fa[35000],f[35000],d[20],c[20];
map<int,string> mapp;
bool vis[20];

void print(int num)
{
    if (!num) return;
    print(num-(1<<(fa[num]-1)));
    cout<<mapp[fa[num]]<<endl;
}

int main()
{
    int t; scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        mapp.clear();
        memset(fa,0,sizeof(fa));
        for (int i=1; i<=n; i++)
        {
            string str; cin>>str;
            mapp[i]=str;
            scanf("%d%d",&d[i],&c[i]);
        }
        memset(f,0x3f3f3f3f,sizeof(f));
        f[0]=0;
        for (int state=0; state<(1<<n); state++)
        {
            memset(vis,false,sizeof(vis));
            int tmp=state;
            int sum=0;
            for (int i=1; i<=n; i++)
            {
                if (tmp&(1<<(i-1)))
                {
                    vis[i]=true;
                    sum+=c[i];
                }
            }
            for (int i=1; i<=n; i++)
            {
                if (vis[i]) continue;
                int tmp;
                if (sum+c[i]<=d[i]) tmp=f[state];
                else tmp=f[state]+sum+c[i]-d[i];
                if (f[state|(1<<(i-1))]>tmp)
                {
                    f[state|(1<<(i-1))]=tmp;
                    fa[state|(1<<(i-1))]=i;
                }
            }
        }
        printf("%d\n",f[(1<<n)-1]);
        print((1<<n)-1);
    }
    return 0;
}