ktv

F - KTV

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

Submit Status

Description

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

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

一直没有看清题意,回溯.

就是找出几个中最大值,对应

a[],b[],c[]

回溯

当step==3,访问过了就退出.

max不能再全局变量中定义

#include <iostream>#include <algorithm>#include <cstring>#include <cstdlib>#include <cstdio>#include <cmath>#include<cstring>using namespace std;struct xys{int x,y,z,s;}a[90];int v[100];int maxx=-1;void dfs(int step,int num,int sum){if(step==3){if( num > maxx )maxx = num;return;}for(int i=0;i<sum;i++){if(  !v[a[i].x] && !v[a[i].y] && !v[a[i].z] ){v[a[i].x] = v[a[i].y] = v[a[i].z] = 1;dfs( step+1, num+a[i].s ,sum);v[a[i].x] = v[a[i].y] = v[a[i].z] = 0;}}}int main(){int time=0;memset(v,0,sizeof(v));int n;while(scanf("%d",&n)){if(n==0)break;     for(int i=0;i<n;i++)    {scanf("%d%d%d%d",&a[i].x,&a[i].y,&a[i].z,&a[i].s);//读数据    }    dfs(0,0,n);        printf("Case %d: %d\n", ++time, maxx );maxx=-1;}}