HDU1074 Doing Home Work - 状压dp

传送门

题目大意：

有n(\(\le 15\))个作业，每个作业有个name, deadline（截止日期），cost（做作业花的时间），如果没有按时完成某个作业，惩罚分数为超出的时间，求一个合理的顺序使得惩罚分数最小，如果有多个方案分数相同，输出字典序最小的。

题目分析

看到\(n \le 15\)可知状压dp：dp[S]表示完成S状态的最小惩罚分数，转移也较为简单：$$dp[i | (1 << (j - 1))] = min(dp[S | (1 << (j - 1))], dp[i] + getTime(i) + cost[j] - deadline[j])$$。因为要输出方案，于是记录一个from数组表示这个状态是从哪个状态转移来的。又因为要求字典序最小，现将name排序，这样在转移时如果\(dp[i | (1 << (j - 1))] == dp[i] + getTime(i) + cost[j] - deadline[j])\)，并且\(i | (1 << (j - 1)) < from[i]\)，那么\(from[i] = i | (1 << (j - 1))\)，来保证字典序最小。

code

#include<bits/stdc++.h>
using namespace std;

const int N = 20, S = 33000, OO = 0x3f3f3f3f;
int n, dp[S], T, from[S];
vector<int> ans;
struct node{
    string name;
    int dl, cost;
    inline bool operator < (const node &b) const {
        return name < b.name;
    }
}task[N];

inline int getTime(int s){
    int ans = 0;
    for (int i=0; (1<<i)<=s; i++) 
        if ((1<<i)&s) 
            ans += task[i + 1].cost;
    return ans;
}

inline void init(){
    memset(dp, OO, sizeof dp), dp[0] = 0;
    memset(from, 0, sizeof from);
    ans.clear();
}

inline int find(int x){
    int l = 1, r = 15;
    while (l<=r) {
        int mid = (l + r) >> 1;
        if (1<<(mid-1) == x) return mid;
        else if (1<<(mid-1) > x) r = mid - 1;
        else l = mid + 1; 
    }
    return 0;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL), cout.tie(NULL);
    cin >> T;
    while(T--){
        init();
        cin >> n;
        for (int i=1; i<=n; i++) cin >> task[i].name >> task[i].dl >> task[i].cost;
        sort(task + 1, task + n + 1);
        for (int i=0; i<(1<<n); i++) {
            for(int j=1; j<=n; j++) {
                if (i&(1<<(j-1))) continue;
                if (dp[i|(1<<(j-1))] > dp[i] + max(0, getTime(i) + task[j].cost - task[j].dl)) {
                    from[i|(1<<(j-1))] = i;
                    dp[i|(1<<(j-1))] = dp[i] + max(0, getTime(i) + task[j].cost - task[j].dl);
                }
                else if(dp[i|(1<<(j-1))] == dp[i] + max(0, getTime(i) + task[j].cost - task[j].dl)) {
                    if(i < from[i|(1<<(j-1))]) from[i|(1<<(j-1))] = i;
                }
            }
        }
        cout << dp[(1<<n)-1] << endl;
        int now = (1<<n)-1;
        while(now) {
            int diff = now ^ (from[now]);
            int pos = find(diff);
            // cout << now << " " << from[now] << " " << diff << " " << pos << endl;
            ans.push_back(pos);
            now = from[now];
        }
        for(int i=ans.size()-1; i>=0; i--) cout << task[ans[i]].name << endl;
    }
    return 0;
}