UVA - 624 - CD （动态规划）

UVA - 624

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Description

You have a long drive by car ahead. You have a tape recorder, but unfortunately your best music is on CDs. You need to have it on tapes so the problem to solve is: you have a tape N minutes long. How to choose tracks from CD to get most out of tape space and have as short unused space as possible.

Assumptions:

• number of tracks on the CD. does not exceed 20
• no track is longer than N minutes
• tracks do not repeat
• length of each track is expressed as an integer number
• N is also integer

Program should find the set of tracks which fills the tape best and print it in the same sequence as the tracks are stored on the CD

Input 

Any number of lines. Each one contains value N, (after space) number of tracks and durations of the tracks. For example from first line in sample data: N=5, number of tracks=3, first track lasts for 1 minute, second one 3 minutes, next one 4 minutes

Output 

Set of tracks (and durations) which are the correct solutions and string `` sum:" and sum of duration times.

Sample Input 

5 3 1 3 4
10 4 9 8 4 2
20 4 10 5 7 4
90 8 10 23 1 2 3 4 5 7
45 8 4 10 44 43 12 9 8 2

Sample Output 

1 4 sum:5
8 2 sum:10
10 5 4 sum:19
10 23 1 2 3 4 5 7 sum:55
4 10 12 9 8 2 sum:45

---

Miguel A. Revilla
2000-01-10

Source

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思路：01背包问题，就是把体积和重量看做了tracks和time，这里注意打印路径的方法

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn= 10005;
int dp[30][maxn], path[30][maxn];
int M, N;
int ti[30];

void print(int a, int b) {
	if(a == 0) return;
	print(a - 1, path[a][b]);
	if(path[a][b] < b) printf("%d ", ti[a]);
}

int main() {
	while(scanf("%d", &M) != EOF) {
		scanf("%d", &N);
		for(int i = 1; i <= N; i++) {
			scanf("%d", &ti[i]);
		}
		
		memset(dp, 0, sizeof(dp));
		for(int i = 1; i <= N; i ++) {
			for(int j = 0; j <= M; j ++) {
				dp[i][j] = dp[i - 1][j];
				path[i][j] = j;//第i个不选时赋值为j;
				if(j >= ti[i]) {
					int tmp = dp[i - 1][j - ti[i]] + ti[i];
					if(tmp > dp[i][j]) {
						dp[i][j] = tmp;
						path[i][j] = j - ti[i];//第i个选了时赋值为j - ti[i]; 
					}
				}
			}
		}
		print(N, M);//递归打印路径 
		printf("sum:%d\n", dp[N][M]);
	}
	return 0;
} 

UVA - 624 - CD （动态规划）

原文：http://blog.csdn.net/u014355480/article/details/44536387