Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) B. Guess the Permutation 水题

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原文链接：http://www.cnblogs.com/qscqesze/p/5179466.html

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#数据结构与算法

本文解析了Codeforces平台上的B.GuessthePermutation题目，介绍了一种通过矩阵还原排列数组的有效方法，并提供了完整的C++代码实现。

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B. Guess the Permutation

题目连接：

http://www.codeforces.com/contest/618/problem/B

Description

Bob has a permutation of integers from 1 to n. Denote this permutation as p. The i-th element of p will be denoted as pi. For all pairs of distinct integers i, j between 1 and n, he wrote the number ai, j = min(pi, pj). He writes ai, i = 0 for all integer i from 1 to n.

Bob gave you all the values of ai, j that he wrote down. Your job is to reconstruct any permutation that could have generated these values. The input will be formed so that it is guaranteed that there is at least one solution that is consistent with the information given.

Input

The first line of the input will contain a single integer n (2 ≤ n ≤ 50).

The next n lines will contain the values of ai, j. The j-th number on the i-th line will represent ai, j. The i-th number on the i-th line will be 0. It's guaranteed that ai, j = aj, i and there is at least one solution consistent with the information given.

Output

Print n space separated integers, which represents a permutation that could have generated these values. If there are multiple possible solutions, print any of them.

Sample Input

2
0 1
1 0

Sample Output

2 1

Hint

题意

给你一个n*n的矩阵，矩阵aij = min(si,sj) aii = 0

然后让你还原s数组，s数组是1-n的排列

题解：

如果某一行出现了n-1个1，那么说明这行一定是1

如果某一行出现了n-2个2，那么这一行一定是2

.....

然后就搞定了，如果某一行出现了n-i个i，那么这一行就是i

代码

#include<bits/stdc++.h>
using namespace std;

int n;
int a[55][55];
int ans[55];
int vis[55];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            scanf("%d",&a[i][j]);
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(vis[j])continue;
            int tot = 0;
            for(int k=1;k<=n;k++)
            {
                if(a[j][k]==i)
                    tot++;
            }
            if(tot==n-i)
            {
                ans[j]=i;
                vis[j]=1;
                break;
            }
        }
    }
    for(int i=1;i<=n;i++)
        printf("%d ",ans[i]);
    printf("\n");
}

转载于:https://www.cnblogs.com/qscqesze/p/5179466.html