【codeforces 814B】An express train to reveries
2017-10-04 18:44
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【题目链接】:http://codeforces.com/contest/814/problem/B
【题意】
给你两个元素个数都为n的序列a[]和b[]
要求你构造出一个排列p(1..n);
使得p与a不同的元素个数恰好为1,p与b不同的元素个数也恰好为1;
保证a和b至少有一个元素不同;
【题解】
稍加分析就能知道;
a和b的不同元素最多只能有两个;
否则无解;
则对不同元素的个数为1个和2个的情况分类讨论就好;
方便起见假设不同的位置在i和j;i<j
个数为1的话;
p[i]与剩下n-1个数字都不一样就好;
个数为2的话;
p[i]=a[i],p[j]=b[i]
或者是p[i]=b[i],p[j] = a[j];
看看哪种符合要求就选那一种.
【Number Of WA】
0
【完整代码】
#include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define LL long long #define rep1(i,a,b) for (int i = a;i <= b;i++) #define rep2(i,a,b) for (int i = a;i >= b;i--) #define mp make_pair #define pb push_back #define fi first #define se second #define ms(x,y) memset(x,y,sizeof x) #define Open() freopen("F:\\rush.txt","r",stdin) #define Close() ios::sync_with_stdio(0),cin.tie(0) typedef pair<int,int> pii; typedef pair<LL,LL> pll; const int dx[9] = {0,1,-1,0,0,-1,-1,1,1}; const int dy[9] = {0,0,0,-1,1,-1,1,-1,1}; const double pi = acos(-1.0); const int N = 1100; int n; int a ,b ,p ,cnt,bo ; vector <int> v; int main(){ //Open(); Close();//scanf,puts,printf not use //init?????? cin >> n; rep1(i,1,n) cin >> a[i]; rep1(i,1,n) cin >> b[i]; rep1(i,1,n) if (a[i]!=b[i]){ cnt++; v.pb(i); } if (cnt==1){ int idx = v[0]; rep1(i,1,n) if (i!=idx){ p[i] = a[i]; bo[a[i]] = 1; } rep1(i,1,n) if (!bo[i]) p[idx] = i; }else{ int idx1 = v[0],idx2 = v[1]; rep1(i,1,n) if (i!=idx1 && i!=idx2){ p[i] = a[i]; bo[a[i]] = 1; } if (!bo[a[idx1]] && !bo[b[idx2]] && a[idx1]!=b[idx2]){ p[idx1] = a[idx1]; p[idx2] = b[idx2]; } if (!bo[b[idx1]] && !bo[a[idx2]] && b[idx1] != a[idx2]){ p[idx1] = b[idx1]; p[idx2] = a[idx2]; } } rep1(i,1,n) cout << p[i] <<' '; return 0; }
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