CodeForces 500B New Year Permutation(序列的变换)
2016-09-22 11:33
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http://codeforces.com/problemset/problem/500/B
B. New Year Permutation
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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)
wherea1 = 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 ifAi, 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 1and 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 thei-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 where1 ≤ 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.
Examples
input
output
input
output
Note
In the first sample, the swap needed to obtain the prettiest permutation is: (p1, p7).
In the second sample, the swaps needed to obtain the prettiest permutation is (p1, p3), (p4, p5), (p3, p4).
A permutation p is a sequence of integers p1, p2, ..., pn,
consisting of n distinct positive integers, each of them doesn't exceed n.
Thei-th element of the permutation p is
denoted as pi.
The size of the permutation p is denoted as n.
新年来临,想要得到比较美丽的序列,
更美丽的序列定义:
序列 a1, a2, ..., an 比序列 b1, b2, ..., bn, 更美丽,当且仅当 k (1 ≤ k ≤ n) 时有 a1 = b1, a2 = b2, ..., ak - 1 = bk - 1 并且 ak < bk 恒成立(题目定义).
可知,更美丽的序列是尽可能的使其按照升序排列。
但是交换序列的条件是,在给定矩阵的行与列值为 1 的时候才可以互换位置。
#include<cstring>
#include<algorithm>
#define AC main()
using namespace std;
const int MYDD = 1103;
int a[MYDD], temp[MYDD], Ans[MYDD];
bool vis[MYDD];//标记 a[] 的使用情况
char Map[MYDD/3][MYDD/3];
int n, it = 0, x;
void DFS(int i) {
if(!vis[a[i]]) x = min(x, a[i]);
temp[i] = it;
for (int j = 0; j < n; j++) {
if (Map[i][j]=='1' && temp[j] != it)
DFS(j);
}
}
int AC {
scanf("%d",&n);
for(int j = 0; j < n; j++)
scanf("%d",&a[j]);
for(int j = 0; j < n; j++)
scanf("%s",&Map[j]);
for(int j = 0; j < n; j++) {
it++, x = n;
DFS(j);
vis[x] = true;
Ans[j] = x;
}
for(int j = 0; j < n; j++) {
printf("%d ", Ans[j]);
}
return 0;
}
B. New Year Permutation
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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)
wherea1 = 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 ifAi, 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 1and 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 thei-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 where1 ≤ 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.
Examples
input
7 5 2 4 3 6 7 1 0001001 0000000 0000010 1000001 0000000 0010000 1001000
output
1 2 4 3 6 7 5
input
5 4 2 1 5 3 00100 00011 10010 01101 01010
output
1 2 3 4 5
Note
In the first sample, the swap needed to obtain the prettiest permutation is: (p1, p7).
In the second sample, the swaps needed to obtain the prettiest permutation is (p1, p3), (p4, p5), (p3, p4).
A permutation p is a sequence of integers p1, p2, ..., pn,
consisting of n distinct positive integers, each of them doesn't exceed n.
Thei-th element of the permutation p is
denoted as pi.
The size of the permutation p is denoted as n.
题意:
新年来临,想要得到比较美丽的序列,更美丽的序列定义:
序列 a1, a2, ..., an 比序列 b1, b2, ..., bn, 更美丽,当且仅当 k (1 ≤ k ≤ n) 时有 a1 = b1, a2 = b2, ..., ak - 1 = bk - 1 并且 ak < bk 恒成立(题目定义).
可知,更美丽的序列是尽可能的使其按照升序排列。
但是交换序列的条件是,在给定矩阵的行与列值为 1 的时候才可以互换位置。
思路:
模拟过一遍。AC CODE:
#include<stdio.h>#include<cstring>
#include<algorithm>
#define AC main()
using namespace std;
const int MYDD = 1103;
int a[MYDD], temp[MYDD], Ans[MYDD];
bool vis[MYDD];//标记 a[] 的使用情况
char Map[MYDD/3][MYDD/3];
int n, it = 0, x;
void DFS(int i) {
if(!vis[a[i]]) x = min(x, a[i]);
temp[i] = it;
for (int j = 0; j < n; j++) {
if (Map[i][j]=='1' && temp[j] != it)
DFS(j);
}
}
int AC {
scanf("%d",&n);
for(int j = 0; j < n; j++)
scanf("%d",&a[j]);
for(int j = 0; j < n; j++)
scanf("%s",&Map[j]);
for(int j = 0; j < n; j++) {
it++, x = n;
DFS(j);
vis[x] = true;
Ans[j] = x;
}
for(int j = 0; j < n; j++) {
printf("%d ", Ans[j]);
}
return 0;
}
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