【CodeForces500B】【贪心】New Year Permutation 题解
来源:互联网 发布:java while和do while 编辑:程序博客网 时间:2024/05/21 04:20
New Year Permutation
User ainta has a permutation p1, p2, …, pn. As the New Year is coming, he wants to make his permutation as pretty as possible.
Permutation a1, a2, …, an is prettier than permutation b1, b2, …, bn, if and only if there exists an integer k (1 ≤ k ≤ n) where a1 = b1, a2 = b2, …, ak - 1 = bk - 1 and ak < bk all holds.
As known, permutation p is so sensitive that it could be only modified by swapping two distinct elements. But swapping two elements is harder than you think. Given an n × n binary matrix A, user ainta can swap the values of pi and pj (1 ≤ i, j ≤ n, i ≠ j) if and only if Ai, j = 1.
Given the permutation p and the matrix A, user ainta wants to know the prettiest permutation that he can obtain.
Input
The first line contains an integer n (1 ≤ n ≤ 300) — the size of the permutation p.
The second line contains n space-separated integers p1, p2, …, pn — the permutation p that user ainta has. Each integer between 1 and n occurs exactly once in the given permutation.
Next n lines describe the matrix A. The i-th line contains n characters ‘0’ or ‘1’ and describes the i-th row of A. The j-th character of the i-th line Ai, j is the element on the intersection of the i-th row and the j-th column of A. It is guaranteed that, for all integers i, j where 1 ≤ i < j ≤ n, Ai, j = Aj, i holds. Also, for all integers i where 1 ≤ i ≤ n, Ai, i = 0 holds.
Output
In the first and only line, print n space-separated integers, describing the prettiest permutation that can be obtained.
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define ms(x, a) memset(a, x, sizeof(a))struct node { int val,pos; } a[1000];inline int cmp(node a, node b) { return a.val < b.val; } const int maxn = 305;int ans[maxn],mp[maxn][maxn];template <class T> inline void read(T &x) { char ch = (char)getchar(); while(ch == ' ' || ch == '\n') ch = (char)getchar(); x = ch-'0';}int main() { int n; scanf("%d",&n); for(int i = 1; i <= n; i++) scanf("%d",&a[i].val), a[i].pos = i; std::sort(a+1, a+n+1, cmp); for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) read(mp[i][j]); for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= n; k++) if(mp[j][i] == 1 && mp[i][k] == 1) mp[j][k] = 1; ms(-1, ans); for(int i = 1; i <= n; i++) { int flag = 0; for(int j = 1; j <= n; j++) if(mp[a[i].pos][j] == 1 && ans[j] == -1) { flag = 1; ans[j] = a[i].val; break; } if(!flag) ans[a[i].pos] = a[i].val; } for(int i = 1; i <= n; i++) printf("%d ",ans[i]); return 0;}
- 【CodeForces500B】【贪心】New Year Permutation 题解
- [CodeForces500B]New Year Permutation[floyd]
- codeforces500b New Year Permutation 【floyd】
- Codeforces 500B New Year Permutation [贪心]
- 【CodeForces 500B】【贪心】New Year Permutation
- CodeForces 500 B. New Year Permutation(贪心+Floyd)
- Codeforces 500B New Year Permutation Floyd算法+贪心
- CF500B New Year Permutation (贪心+并查集)
- B - New Year Permutation
- B. New Year Permutation
- CF_500B New Year Permutation
- New Year Permutation
- New Year Permutation
- New Year Permutation
- New Year Permutation
- CodeForces 500B New Year Permutation (Floyd判断通路是否存在)(贪心)
- 【Goodbye2014】Codeforces 500B New Year Permutation【贪心+Floyd传递闭包】
- codeforces New Year Present 题解
- pat 甲级 1104. Sum of Number Segments
- spring DI起步
- 大数据-二
- 神经网络学习笔记(二):feedforward和feedback
- Android 保存图片到SQLite,读出SQLite中的图片
- 【CodeForces500B】【贪心】New Year Permutation 题解
- Mybatis知识点备忘
- Spring事务的隔离级别
- 引用数组,指针数组与数组引用,数组指针
- ffmpeg(二) 网络流转发
- 寻找两个链表的第一个公共结点
- 写个swiper小demo
- SqlHelper
- linux init.rc service name 过长导致服务不可用的问题