5821: Clothes（暴力？）

最新推荐文章于 2020-01-13 23:19:39 发布

原创最新推荐文章于 2020-01-13 23:19:39 发布 · 412 阅读

1 ·

CC 4.0 BY-SA版权

***ACM***思维*** 专栏收录该内容

42 篇文章

订阅专栏

本题描述了一个小男孩Gerald面临的问题：如何从一家售卖多种衣物的商店中选择三件相互搭配的衣物，并且使得花费最少。输入包括衣物总数、可搭配组合数及各衣物价格等信息。

5821: Clothes

时间限制: 2 Sec 内存限制: 256 MB
提交: 235 解决: 49
[ 提交][ 状态][ 讨论版][命题人: admin]

题目描述

A little boy Gerald entered a clothes shop and found out something very unpleasant: not all clothes turns out to match. For example, Gerald noticed that he looks rather ridiculous in a smoking suit and a baseball cap.

Overall the shop sells n clothing items, and exactly m pairs of clothing items match. Each item has its price, represented by an integer number of rubles. Gerald wants to buy three clothing items so that they matched each other. Besides, he wants to spend as little money as possible. Find the least possible sum he can spend.

输入

The first input file line contains integers n and m — the total number of clothing items in the shop and the total number of matching pairs of clothing items (3≤n≤100,0≤m≤n(n-1)/2).

Next line contains n integers ai (1≤ai≤106) — the prices of the clothing items in rubles.

Next m lines each contain a pair of space-separated integers ui and vi (1≤ui,vi≤n,ui≠vi). Each such pair of numbers means that the ui-th and the vi-th clothing items match each other. It is guaranteed that in each pair ui and vi are distinct and all the unordered pairs (ui,vi) are different.

输出

Print the only number — the least possible sum in rubles that Gerald will have to pay in the shop. If the shop has no three clothing items that would match each other, print "-1" (without the quotes).

样例输入

样例输出

提示

In the first test there only are three pieces of clothing and they all match each other. Thus, there is only one way — to buy the 3 pieces of clothing; in this case he spends 6 roubles.

来源

#include<stdio.h>
#include<string.h>
#define inf 0x3f3f3f3f
#include<algorithm>
using namespace std;
 
int main()
{
    int m, n, a[105], minn=9999999, map[105][105], x, y, ans;
    scanf("%d%d", &n, &m);
    for(int i=1;i<=n;i++)
        scanf("%d", &a[i]);
    memset(map, 0, sizeof(map));
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d", &x, &y);
        map[x][y]=map[y][x]=1;
    }
    for(int k=1;k<=n;k++)
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                if(map[i][k] && map[k][j] && map[i][j])
                {
                    ans=a[i]+a[j]+a[k];
                    if(ans<minn)
                        minn=ans;
                }
    if(minn==9999999)
        printf("-1\n");
    else
        printf("%d\n", minn);
    return 0;
}
 
/**************************************************************
    Problem: 5821
    User: ldu_reserver201701
    Language: C++
    Result: 正确
    Time:68 ms
    Memory:1092 kb
****************************************************************/