HDU 1074

http://acm.hdu.edu.cn/showproblem.php?pid=1074

状压dp，记录路径

求最小值的状压dp非常裸，5分钟就写好了，记录路径有点麻烦，之前没怎么处理过这种问题

我的方法是用一个map建立当前状态和前驱状态的映射，输出要按字典序，因为已经按字典序从大到小排好了，所以状态值比较小的优先（字典序小）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
using namespace std;

const int INF = 0xfffffff;

struct fuck{
    int w, pre;
}dp[1<<15];

struct node{
    char name[105];
    int D, C;
}p[16];

int cal(int x){
    int res = 0, cnt = 0;
    while(x){
        if(x & 1){
            res += p[cnt].C;
        }
        cnt++;
        x >>= 1;
    }
    return res;
}

int cnt(int x){
    int res = 0;
    while(x){
        if(x & 1){
            res++;
        }
        x >>= 1;
    }
    return res;
}

int cal1(int x, int y){
    int res = 0;
    while(1){
        if((x&1) != (y&1)){
            return res; 
        }
        res++; 
        x >>= 1; y >>= 1; 
    }
}

int ans[25], ct;
    
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 0; i < (1<<n); i++)
            dp[i].w = INF;
        int xx = 0;
        while(1){
            dp[(1<<xx)].pre = 0; 
            xx++; 
            if(xx == n) break;
        }
        for(int i = 0;i < n; i++){
            scanf("%s %d %d", p[i].name, &p[i].D, &p[i].C);
            if(p[i].C > p[i].D)
                dp[1<<i].w = p[i].C - p[i].D;
            else 
                dp[1<<i].w = 0;
        }
        map <int, int> mp; 
        for(int i = 0; i < (1<<n); i++){
            for(int j = 0; j < n; j++){
                if(i&(1<<j)){
                    if(cal(i) > p[j].D){
                        if(dp[i].w > dp[i^(1<<j)].w + cal(i) - p[j].D){
                            dp[i].w = dp[i^(1<<j)].w + cal(i) - p[j].D;
                            dp[i].pre = i^(1<<j); 
                            mp[i] = i^(1<<j);    
                        }
                        else if(dp[i].w == dp[i^(1<<j)].w + cal(i) - p[j].D && dp[i].pre > (i^(1<<j))){
                            dp[i].pre = i^(1<<j); 
                            mp[i] = i^(1<<j);      
                        }
                            //dp[i] = min(dp[i], dp[i^(1<<j)] + cal(i) - p[j].D);
                    }
                    else{
                        if(dp[i].w > dp[i^(1<<j)].w){
                            dp[i].w = dp[i^(1<<j)].w;
                            dp[i].pre = i^(1<<j);
                            mp[i] = i^(1<<j); 
                        }
                        else if(dp[i].w == dp[i^(1<<j)].w && dp[i].pre > (i^(1<<j))){
                            dp[i].pre = i^(1<<j);
                            mp[i] = i^(1<<j); 
                        }
                        //dp[i] = min(dp[i], dp[i^(1<<j)]);
                    }
                }
            }
        }
        printf("%d\n", dp[(1<<n)-1]);
        int now = (1<<n) - 1; 
        ct = 0; 
        while(now){
            ans[ct++] = cal1(now, mp[now]);
            now = mp[now]; 
        }
        for(int i = n - 1; i >= 0; i--){
            printf("%s\n", p[ans[i]].name);
        }
    }
    return 0;
}