codeforces B. Shower Line 解题报告

题目链接：http://codeforces.com/contest/431/problem/B

题目意思：给出5 * 5 的矩阵。从这个矩阵中选出合理的安排次序，使得happiness之和最大。当第i个人和第j个人talk 的时候，第i个人获得的happiness是g[i][j]，第j 个人获得的happiness是g[j][i]。

    好简单的一道题目，不知道昨晚徘徊好久都不敢打，五个for循环即可！数据量这么小......今天一次就过了...

    谨以此来纪念自己的怯懦....

    方法一：直接暴力枚举

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define LL long long
int g[10][10];
LL ans;

int main()
{
    for (int i = 1; i <= 5; i++)
    {
        for (int j = 1; j <= 5; j++)
            scanf("%d", &g[i][j]);
    }
    ans = 0;
    for (int i = 1; i <= 5; i++)
    {
        for (int j = 1; j <= 5; j++)
        {
            if (i != j)
            {
                for (int k = 1; k <= 5; k++)
                {
                    if (k != j && k != i)
                    {
                        for (int l = 1; l <= 5; l++)
                        {
                            if (l != k && l != i && l != j)
                            {
                                for (int p = 1; p <= 5; p++)
                                {
                                    if (p != l && p != i && p != j && p != k)
                                    {
                                 //       printf("i = %d, j = %d, k = %d, l = %d, p = %d
", i, j, k, l, p);
                                        LL sum = g[i][j] + g[j][i] + g[j][k] + g[k][j] + 2 * (g[l][p] + g[p][l] + g[k][l] + g[l][k]);
                                        ans = max(ans, sum);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    printf("%lld
", ans);
    return 0;
}

     方法二：利用next_permutation （学人家代码写的）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std;

#define LL long long
const int maxn = 5;
int grid[maxn+1][maxn+1];
int t[maxn];
LL ans, tmp;

int main()
{
    for (int i = 0; i < maxn; i++)
    {
        for (int j = 0; j < maxn; j++)
            scanf("%d", &grid[i][j]);
    }
    for (int i = 0; i < maxn; i++)
        t[i] = i;
    ans = 0;
    do
    {
        //0123: 01 talk  23 talk
        tmp = grid[t[0]][t[1]] + grid[t[1]][t[0]];
        tmp += grid[t[2]][t[3]] + grid[t[3]][t[2]];
        //1234: 12 talk  34 talk
        tmp += grid[t[1]][t[2]] + grid[t[2]][t[1]];
        tmp += grid[t[3]][t[4]] + grid[t[4]][t[3]];
        //23: 23 talk
        tmp += grid[t[2]][t[3]] + grid[t[3]][t[2]];
        //34: 34 talk
        tmp += grid[t[3]][t[4]] + grid[t[4]][t[3]];

        ans = max(ans, tmp);
    }while (next_permutation(t, t+maxn));
    printf("%lld
", ans);
    return 0;
}