UVA 473 - Raucous Rockers(dp)

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 Raucous Rockers  You just inherited the rights to n previously unreleased songs recorded by the popular group Raucous Rockers. You plan to release a set of m compact disks with a selection of these songs. Each disk can hold a maximum of t minutes of music, and a song can not overlap from one disk to another. Since you are a classical music fan and have no way to judge the artistic merits of these songs, you decide on the following criteria for making the selection: The songs will be recorded on the set of disks in the order of the dates they were written.The total number of songs included will be maximized. Input The input consists of several datasets. The first line of the input indicates the number of datasets, then there is a blank line and the datasets separated by a blank line. Each dataset consists of a line containing the values of n, t and m (integer numbers) followed by a line containing a list of the length of n songs, ordered by the date they were written (Each  is between 1 and t minutes long, both inclusive, and  .) Output The output for each dataset consists of one integer indicating the number of songs that, following the above selection criteria will fit on m disks. Print a blank line between consecutive datasets. Sample Input 2 10 5 3 3, 5, 1, 2, 3, 5, 4, 1, 1, 5 1 1 1 1 Sample Output 6 1 题意:有n首音乐,m个磁盘,磁盘容量都是t,一首音乐不能被分割到两个盘里,问最多能在磁盘里放置多少音乐。要求音乐的放置不能逆序。即下标大的不能放在下标小的前面。 思路:dp[i][j][k]表示第i首音乐,放在第j个磁盘的k位置,这样就看磁盘放不放,如果不放dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j][k]);如果放,如果是放在第一个位置的情况要特殊考虑,因为放在第一个位置的话等于上一张磁盘的最后一个位置dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j - 1][t] + 1);否则为dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j][k - cd[i]] + 1);然后可以利用滚动数组去优化,少掉第一维。 代码: #include <stdio.h> #include <string.h> #define max(a,b) ((a)>(b)?(a):(b)) const int N = 105; int T, n, t, m, cd, dp[N][N]; int solve() { memset(dp, 0, sizeof(dp)); for (int i = 1; i <= n; i++) { scanf("%d%*c", &cd); for (int j = m; j >= 1; j--) { for (int k = t; k >= 1; k--) { if (k < cd) continue; dp[j][k] = max(dp[j][k], dp[j - 1][t] + 1); dp[j][k] = max(dp[j][k], dp[j][k - cd] + 1); } } } return dp[m][t]; } int main() { scanf("%d", &T); while (T--) { scanf("%d%d%d", &n, &t, &m); printf("%d\n", solve()); if (T) printf("\n"); } return 0; }