http://www.elijahqi.win/archives/737
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.
这题我们做的时候可以考虑把题中所有能交换位置的点用并查集当连通块连接起来，然后针对并查集内元素进行排序
贪心求最小 然后把这个集合中能摆放的位置也从小到大摆放，然后一一对应（贪心策略）
#include<cstdio>#include<vector>
#include<algorithm>
#define N 330
using namespace std;
inline int read(){
int x=0;char ch=getchar();
while (ch<'0'||ch>'9') ch=getchar();
while (ch<='9'&&ch>='0') {x=x*10+ch-'0';ch=getchar();}
return x;
}
int fa[N],a[N],ans[N];
inline int find(int x){
return x==fa[x]?x:fa[x]=find(fa[x]);
}
vector<int> p[N];
void solve(int x){
vector<int> tmp;
for (int i=0;i<p[x].size();++i) tmp.push_back(a[p[x][i]]);
//while (tmp.size()) printf("%d ",tmp.back()),tmp.pop_back();printf("\n");
sort(tmp.begin(),tmp.end());sort(p[x].begin(),p[x].end());
for (int i=0;i<tmp.size();++i) ans[p[x][i]]=tmp[i];
}
int n;
int main(){
freopen("cf.in","r",stdin);
n=read();char str1[N];
for (int i=1;i<=n;++i) a[i]=read(),fa[i]=i;
for (int i=1;i<=n;++i){
scanf("%s",str1+1);
for (int j=1;j<=n;++j) if (str1[j]=='1'){
int x=find(i),y=find(j);
if(x!=y) fa[x]=y;
}
}
for (int i=1;i<=n;++i) p[find(i)].push_back(i);
for (int i=1;i<=n;++i) if (p[i].size()) solve(i);
for (int i=1;i<=n;++i) printf("%d ",ans[i]);
return 0;
}