hdu4833 Best Financing(DP)

  题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4833

  这道题目关键的思想是从后往前dp,dp[i]表示在第i处投资xi能获得的最大收益,其中xi表示从第i-1到第i处投资之间的收入总和(不包括收益),也就是用xi这些钱从第i处开始投资到最后靠xi这些钱获得的最大收益。那么获得的总收益就是dp[0] * x0 + dp[1] * x1 + ... + dp[n] * xn。另一个关键思想就是对理财产品的起始和结束时间进行离散化,这样能够节约空间和dp的时间。

 1 #include <stdio.h>
 2 #include <vector>
 3 #include <map>
 4 #include <string.h>
 5 #include <algorithm>
 6 using namespace std;
 7 
 8 int dates[100005], ls[2505 * 2], jg[2505 * 2];
 9 long long dp[2505 * 2], ans;
10 struct Product
11 {
12     int start, finish, rate;
13 }product[2505];
14 map<int, int> myMap;
15 vector<int> vec1[2505 * 2], vec2[2505 * 2];
16 
17 #define max(a, b) ((a) > (b) ? (a) : (b))
18 
19 int main(void)
20 {
21     int n, m, t, i, j, k, date, earning, cnt, maxDate;
22 
23     scanf("%d", &t);
24     for (i = 1; i <= t; i++)
25     {
26         scanf("%d%d", &n, &m);    
27         memset(dates, 0, sizeof(dates));
28         maxDate = 0;
29         for (j = 0; j < n; j++)
30         {
31             scanf("%d%d", &date, &earning);    
32             dates[date] += earning;
33             if (date > maxDate)
34                 maxDate = date;
35         }
36         for (j = 0; j < m; j++)
37         {
38             scanf("%d%d%d", &product[j].start, &product[j].finish, &product[j].rate);
39             ls[j] = product[j].start;
40             ls[j + m] = product[j].finish;
41             if (product[j].finish > maxDate)
42                 maxDate = product[j].finish;
43         }
44         for (j = 1; j <= maxDate; j++)
45             dates[j] += dates[j - 1];
46         sort(ls, ls + m + m, less<int>());
47         cnt = unique(ls, ls + m + m) - ls;    
48         jg[0] = dates[ls[0]]; 
49         myMap.clear();
50         for (j = 0; j < cnt; j++)
51         {
52             myMap[ls[j]] = j;
53             jg[j] = dates[ls[j]] - dates[ls[j - 1]];
54             vec1[j].clear();
55             vec2[j].clear();
56         }
57         for (j = 0; j < m; j++)
58         {
59             vec1[myMap[product[j].start]].push_back(myMap[product[j].finish]);
60             vec2[myMap[product[j].start]].push_back(product[j].rate);
61         }
62         memset(dp, 0, sizeof(dp));
63         for (j = cnt - 1; j >= 0; j--)
64         {
65             dp[j] = dp[j + 1];
66             for (k = 0; k < vec1[j].size(); k++)
67                 dp[j] = max(dp[j], dp[vec1[j][k]] + vec2[j][k]);
68         }
69         ans = 0;
70         for (j = 0; j < cnt; j++)
71             ans += dp[j] * jg[j];
72         printf("Case #%d:\n%.2lf\n", i, (double)ans / 100.0);
73     }
74     return 0;
75 }

你可能感兴趣的:(HDU)