UVA 11218 (13.11.09)

最新推荐文章于 2021-11-21 11:09:01 发布

最新推荐文章于 2021-11-21 11:09:01 发布 · 127 阅读

本文介绍了一种解决KTV唱歌分组问题的算法。在9个人中选出3人一组的最优组合，使得每组得分最高。通过回溯算法遍历所有可能的组合并找出最佳解决方案。

Problem K

KTV

One song is extremely popular recently, so you and your friends decided to sing it in KTV. The song has 3 characters, so exactly 3 people should sing together each time (yes, there are 3 microphones in the room). There are exactly 9 people, so you decided that each person sings exactly once. In other words, all the people are divided into 3 disjoint groups, so that every person is in exactly one group.

However, some people don't want to sing with some other people, and some combinations perform worse than others combinations. Given a score for every possible combination of 3 people, what is the largest possible score for all the 3 groups?

Input

The input consists of at most 1000 test cases. Each case begins with a line containing a single integer n (0 < n < 81), the number of possible combinations. The next n lines each contains 4 positive integers a, b, c, s (1 <= a < b < c <= 9, 0 < s < 10000), that means a score of s is given to the combination (a,b,c). The last case is followed by a single zero, which should not be processed.

Output

For each test case, print the case number and the largest score. If it is impossible, print -1.

Sample Input

Output for the Sample Input

Case 1: 6
Case 2: -1

题意：有那么一首歌，需要三个人一起唱。现在KTV里有9个人了，显然分成三组唱，但问题是，在给出的分组方案中怎么选择，可以使三组总得分最高(注意: 每个人都要唱过!!)

做题总结：做这道题让我对回溯加深了印象，其实回溯就是深搜，但是为什么要回溯呢，因为当前这个组可能要，也可能不要，所以每次把当前组搜一下就回来，并把标记改回去，这就是这道题的重点了。

具体看一下AC代码：

#include<stdio.h>
#include<string.h>

int n;
int max;
int vis[10];

struct Group {
    int a;
    int b;
    int c;
    int s;
} g[100];

void dfs(int num, int score) {
    if(num == 3) {
        if(score > max)
            max = score;
        return ;
    }
    for(int i = 0; i < n; i++) {
        if(!vis[g[i].a] && !vis[g[i].b] && !vis[g[i].c]) {
            vis[g[i].a] = vis[g[i].b] = vis[g[i].c] = 1;
            dfs(num + 1, score + g[i].s);
            vis[g[i].a] = vis[g[i].b] = vis[g[i].c] = 0;
        }
    }
}

int main() {
    int cse = 0;
    while(scanf("%d", &n) != EOF, n) {
        for(int i = 0; i < n; i++)
            scanf("%d %d %d %d", &g[i].a, &g[i].b, &g[i].c, &g[i].s);
        memset(vis, 0, sizeof(vis));
        max = 0;
        dfs(0, 0);
        if(max)
            printf("Case %d: %d\n", ++cse, max);
        else
            printf("Case %d: -1\n", ++cse);
    }
    return 0;
}