CodeFroces 814B An express train to reveries(构造题)
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http://codeforces.com/problemset/problem/814/B
题意中,有n个流星,然后看见了两次它们的颜色,每次都只有一个流星的颜色是错误的,题目保证答案存在,输出一种可行的答案。
解法,显然,如果第i个位置,a和b相同,则这个位置肯定是正确的。所以我们只需记录下不相同的位置的坐标,然后暴力。
样例中也给出了,如果只有一个位置是不相同的,那么肯定这个位置a和b的颜色都是错误的,所以我们要找出1到n哪个数字没有用过。
如果有两个位置是不相同的,那么组合也就两种,直接暴力,判断一下是否符合,不符合换另一种。
代码如下:
#include<bits/stdc++.h>using namespace std;int main(){int n, a[1005], b[1005], ans[1005] = {0}, key[5], k = 0;cin >> n;for(int i = 0; i < n; i++)scanf("%d", &a[i]);for(int i = 0; i < n; i++)scanf("%d", &b[i]);for(int i = 0; i < n; i++){if(a[i] == b[i])ans[i] = a[i];elsekey[k++] = i;}priority_queue <int> q;if(k == 1){for(int i = 0; i < n; i++)if(i != key[0])q.push(ans[i]);for(int i = n; i > 0; i--){if(q.top() != i){ans[key[0]] = i;break;}q.pop();}for(int i = 0; i < n; i++)printf("%d%c", ans[i], i == n - 1 ? '\n' : ' ');}else if(k == 2){ans[key[0]] = b[key[0]];ans[key[1]] = a[key[1]];for(int i = 0; i < n; i++){q.push(ans[i]);}bool flag = 0;for(int i = n; i > 0; i--){if(q.top() != i){flag = 1;break;}q.pop();}if(flag){ans[key[0]] = a[key[0]];ans[key[1]] = b[key[1]];}for(int i = 0; i < n; i++)printf("%d%c", ans[i], i == n - 1 ? '\n' : ' ');}return 0;}
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