UVa 11218 – KTV， Rujia Liu的神題（一）

連結：

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

類型：   暴力回溯

原題：

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

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

題目大意：

有9個人去KTV唱歌， 然後要分組一起唱歌，每組3人，一人只能分在一個組裡。  然而不同的組合效果不同， 而且某些人不想跟某些人同一組。 所以不同的組合的得分是不同的。求出這些組合中最高能得到的總分是多少。

分析與總結：

這題是Rujia
Liu's Problems for Beginners專題的第一題， 也是裡面最簡單的一題。

只需要暴力的回溯枚舉出合格情況，取其中最高的分數及可。

/* *  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;}

——  生命的意義，在於賦予它意義。
               原創 http://blog.csdn.net/shuangde800 ， By   D_Double  (轉載請標明)