【HDU 1074】Doing Homework（状压+记忆化）

DFS策略：对于当前状态，枚举下一个选哪个(i)，如果当前耗时已经超过i的死亡线，ans就累积

然后可以记忆化当前状态标记为true 之后扫到这儿就return

代码不是A的。。和几份题解都对拍了半天没找出问题 如果能指出错误的话请评论

#include<bits/stdc++.h>
#define N 17
#define INF 0x3f3f3f3f
using namespace std;
int T,n,a[N],b[N],ans;
bool vis[1<<N];
struct Data
{
    string name;
    int dead,tim;
}sub[N];
inline int count(int x)
{
    int num=0;
    while(x)
    {
        if(x&1) num++;
        x>>=1;
    }
    return num;
}
void dfs(int state,int day,int reduce)
{
    if(state==(1<<n)-1)
    {
        if(ans>reduce)
        {
            ans=reduce;
            for(int i=1;i<=n;i++)    b[i]=a[i];
        }
        return;
    }
    if(vis[state])    return; 
    vis[state]=true;
    for(int i=1;i<=n;i++)
    {
        if(state&(1<<(i-1)))    continue;
        int num=count(state)+1;    //第几个选的 
        a[num]=i;
        if(day+sub[i].tim>sub[i].dead)    //如果加上i的天数已经超过i的死亡线 
            dfs(state|(1<<(i-1)),day+sub[i].tim,reduce+day+sub[i].tim-sub[i].dead); 
        else dfs(state|(1<<(i-1)),day+sub[i].tim,reduce);
        a[num]=0;
    }
}
int main()
{
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=1;i<=n;i++)
        {
            cin>>sub[i].name>>sub[i].dead>>sub[i].tim;    
        }
        ans=INF;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(vis,false,sizeof(vis));
        dfs(0,0,0);
        cout<<ans<<endl;
        for(int i=1;i<=n;i++)    cout<<sub[b[i]].name<<endl;
    }
    return 0;
}