UVa 11218 - KTV， Rujia Liu的神人题（一）

UVa 11218 - KTV， Rujia Liu的神题（一）

链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=112&page=show_problem&problem=2159

类型：   暴力回溯

原题：

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?

InputThe 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.

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

Sample Input

31 2 3 14 5 6 27 8 9 341 2 3 11 4 5 21 6 7 31 8 9 40

Output for the Sample Input

Case 1: 6Case 2: -1

题目大意：

有9个人去KTV唱歌， 然后要分组一起唱歌，每组3人，一人只能分在一个组里。  然而不同的组合效果不同， 而且某些人不想跟某些人同一组。 所以不同的组合的得分是不同的。求出这些组合中最高能得到的总分是多少。

分析与总结：

这题是Rujia Liu's Problems for Beginners专题的第一题， 也是里面最简单的一题。

只需要暴力的回溯枚举出符合条件的情况，取其中最高的分数及可。

/* *  UVa  11218 - KTV *   回溯 *  Time: 0.312s (UVa) *  Author: D_Double */#include<iostream>#include<cstdio>#include<cstring>using namespace std;int group[82][4], ans, n;bool vis[82], occur[10];void search(int cur,int tot){    if(cur >= 3){        if(tot > ans) ans = tot;        return;    }    for(int i=0; i<n; ++i)if(!vis[i]){        if(occur[group[i][0]] || occur[group[i][1]] || occur[group[i][2]])            continue;        occur[group[i][0]] = occur[group[i][1]] = occur[group[i][2]] = true;        vis[i] = true;        search(cur+1, tot+group[i][3]);        occur[group[i][0]] = occur[group[i][1]] = occur[group[i][2]] = false;        vis[i] = false;    } }int main(){    int cas=1;    while(scanf("%d", &n), n){        for(int i=0; i<n; ++i)            scanf("%d%d%d%d",&group[i][0],&group[i][1],&group[i][2],&group[i][3]);        ans = -1;        memset(vis, 0, sizeof(vis));        memset(occur, 0, sizeof(occur));        search(0, 0);        printf("Case %d: %d\n", cas++, ans);    }    return 0;}

——  生命的意义，在于赋予它意义。
               原创 http://blog.csdn.net/shuangde800 ， By   D_Double  (转载请标明)