标签：acm   状态压缩   

题意  有n个作业要做   给你每个作业的最后期限  和做完这个作业需要的时间  作业每超过最后期限一天就会扣一分  只能把一个作业做完了再做另一个作业  问做完所有作业至少扣多少分

作业最多只有15个  看到这个数字容易想到是状态压缩  dp[i]表示i对应状态的最小扣分  i转换为二进制后为1的位表明该位对应的作业已经做了  为0的位没做  那么dp[i] = min{dp[k] + cost | k为将某一位变成1后等于 i 的状态} 

由于要打印路径  所有还需要记录每个状态的上一个状态pre

#include <cstdio>
#include <cstring>
using namespace std;
const int N = 16, M = 1 << N;
int d[N], c[N], n;
int dp[M], pre[M], t[M];
char s[N][105];

void print(int k)
{
    if(k == 0) return;
    print(pre[k]);
    k -= pre[k];
    for(int i = 0; i < n; ++i)
        if(k & 1 << i) puts(s[i]);
}

int main()
{
    int T, m, cost;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i = 0; i < n; ++i)
            scanf("%s%d%d", s[i], &d[i], &c[i]);
        memset(dp, 0x3f, sizeof(dp));
        dp[0] = t[0] = 0;  //边界 所有作业都没做的扣分为0
        m = 1 << n;

        for(int i = 1; i < m; ++i)
        {
            for(int j = 0; j < n; ++j)
            {
                if((i & 1 << j) == 0) continue;
                int k = i - (1 << j);
                t[i] = t[k] + c[j];
                cost = t[i] > d[j] ? t[i] - d[j] : 0;
                if(dp[k] + cost <= dp[i])  //'='保证字典序最小
                {
                    dp[i] = dp[k] + cost;
                    pre[i] =  k;  //记录路径
                }
            }
        }
        printf("%d\n", dp[m - 1]);
        print(m - 1);  //打印路径
    }
    return 0;
}

Doing Homework

2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3

2
Computer
Math
English
3
Computer
English
Math

Hint
In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the 
word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.
 

HDU 1074 Doing Homework（DP·状态压缩）

标签：acm   状态压缩   

原文地址：http://blog.csdn.net/acvay/article/details/48051975