USACO: Raucous Rockers

Raucous Rockers

You just inherited the rights to N (1 <= N <= 20) previously unreleased songs recorded by the popular group Raucous Rockers. You plan to release a set of M (1 <= M <= 20) compact disks with a selection of these songs. Each disk can hold a maximum of T (1 <= T <= 20) 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 on the set of disks must appear in the order of the dates that they were written.
• The total number of songs included will be maximized.

PROGRAM NAME: rockers

INPUT FORMAT

Line 1:	Three integers: N, T, and M.
Line 2:	N integers that are the lengths of the songs ordered by the date they were written.

SAMPLE INPUT (file rockers.in)

4 5 2
4 3 4 2

OUTPUT FORMAT

A single line with an integer that is the number of songs that will fit on M disks.

SAMPLE OUTPUT (file rockers.out)

3

 

解题思路：

DP，不过状态检测转移量比较大，我的这个算法大概是θ((n*m)^2)，好在题目给的数据量不大

DP的状态转移参见代码注释：其实DP原理还是要自己想得通才行

注意我定义的best[i][j]是：i张碟，第j首歌添加在最后那张碟上，此时可以获得的最多曲目数；如果不能达到这个要求，比如第j首太长根本无法刻入，或者其实第j首可以刻到倒数第二张碟上，best[i][j]都被设置为-1

/* ID: geterns1 PROG: rockers LANG: C++ */ #include <iostream> #include <fstream> #include <cstdio> #include <cmath> #include <cstdlib> #include <cstring> #include <algorithm> using namespace std; int main(){ freopen("rockers.in", "r", stdin); freopen("rockers.out", "w", stdout); // int n, t, m, song[20]; scanf("%d %d %d", &n, &t, &m); for (int i = 0; i < n; i ++) scanf("%d", song + i); /* best[i][j] keep the maximum number of songs with i disks, last added j-th song (and, exactly, the j-th song must be added at the i-th disk, otherwise, set a -1) remi[i][j] keep the corresponding maximum remainder time of the last disk */ int best[21][21], remi[21][21]; //initialization best[0][0] = 0; remi[0][0] = 0; for (int i = 1; i <= n; i ++){ best[0][i] = -1; remi[0][i] = 0; } for (int i = 1; i <= m; i ++){ best[i][0] = -1; remi[i][0] = 0; } //DP int b, r; for (int i = 1; i <= m; i ++){ for (int j = 1; j <= n; j ++){ //noway to add this song if (song[j - 1] > t){ best[i][j] = -1; continue; } //state transition: add a disk or add on the last disk b = -1; r = 0; //for best[i][j], check best[i - 1][0]~best[i - 1][j - 1] for (int k = 0; k < j; k ++){ if (best[i - 1][k] > -1 && remi[i - 1][k] < song[j - 1]){ if (best[i - 1][k] + 1 >= b){ if (best[i - 1][k] + 1 > b){ b = best[i - 1][k] + 1; r = t - song[j - 1]; } else if (t - song[j - 1] > r) r = t - song[j - 1]; } } } //for best[i][j], check best[i][0]~best[i][j - 1] for (int k = 0; k < j; k ++){ if (best[i][k] > -1 && remi[i][k] >= song[j - 1]){ if (best[i][k] + 1 >= b){ if (best[i][k] + 1 > b){ b = best[i][k] + 1; r = remi[i][k] - song[j - 1]; } else if (remi[i][k] - song[j - 1] > r) r = remi[i][k] - song[j - 1]; } } } best[i][j] = b; remi[i][j] = r; } } // int MAX = 0; for (int i = 1; i <= m; i ++) for (int j = 1; j <= n; j ++) if (best[i][j] > MAX) MAX = best[i][j]; printf("%d/n", MAX); fclose(stdin); fclose(stdout); return 0; }

 

 

顺便贴一下官方的题解

 /* This is a pretty straight-forward dynamic programming problem. The factors that determine whether we can put a song on a CD are: The length used up on the CD so far The length of the current song The last song that we put on the CD (because of the date restriction) Therefore, we create an matrix called "dp", with dp[a][b][c] being the most number of songs that we can put on the first 'a' CDs, with 'b' minutes already used up on the 'a'th CD, and with 'c' being the last song that we put on CD 'a'. We initialize the matrix to be all zeroes, and then we cycle through it as follows: We traverse the CDs in ascending order to satisfy the date requirement We go through the number of minutes used in ascending order, so that we have had a chance to put on a song earlier in the CD before we try to put songs on at a later time We go through the last song used in ascending order (although this order doesn't really matter) We go through the new songs in ascending order, making sure to start with songs that were dated after the last song If the new song that we want to include in the set will fit on the current CD, there is no reason to put it on the next CD, so we check the matrix and see if adding it to the current set of songs will be better than the value already stored in the matrix. Otherwise, we check to see if it would be beneficial to put the song on the next CD. As a time-saver, each time we check a current value in the matrix to see what other songs we can put on the CD, we also check and see if this value is better than the most number of CDs that we have currently been able to fit in the set, so that we don't need to do this at the end. Finally, we output this best number, which will be the most number of CDs that we can fit in the set. */ #include <stdio.h> #define MAX 25 int dp[MAX][MAX][MAX], length[MAX]; int main () { FILE *in = fopen ("rockers.in", "r"); FILE *out = fopen ("rockers.out", "w"); int a, b, c, d, best, numsongs, cdlength, numcds; fscanf (in, "%d%d%d", &numsongs, &cdlength, &numcds); for (a = 1; a <= numsongs; a++) fscanf (in, "%d", &length[a]); best = 0; for (a = 0; a < numcds; a++)/* current cd */ for (b = 0; b <= cdlength; b++) /* number of minutes used */ for (c = 0; c <= numsongs; c++) { /* last song */ for (d = c + 1; d <= numsongs; d++) { /* new song */ if (b + length[d] <= cdlength) { if (dp[a][b][c] + 1 > dp[a][b + length[d]][d]) dp[a][b + length[d]][d] = dp[a][b][c] + 1; } else { if (dp[a][b][c] + 1 > dp[a + 1][length[d]][d]) dp[a + 1][length[d]][d] = dp[a][b][c] + 1; } } if (dp[a][b][c] > best) best = dp[a][b][c]; } fprintf (out, "%d/n", best); return 0; }