題目鏈接

BZOJ4104

題解

我們將已知字符排序，由循環就可以得到一個對應關系
如樣例就是：
0->第5行
1->第1行
1->第2行
1->第3行
1->第5行
2>第6行
3->第4行
按照這個循序加入答案即可

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
 

#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 200005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while
 
 (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
int n,m,A[maxn],B[maxn],ans[maxn];
cp lk[maxn];
int main(){
    n = read(); m = read();
    REP(i,n + 1
 
) A[i] = B[i] = read();
    sort(B + 1,B + 1 + n + 1);
    REP(i,n + 1) lk[i] = mp(A[i],i);
    sort(lk + 1,lk + 1 + n + 1);
    for (int now = 1,i = 1; i <= n; i++,now = lk[now].second){
        ans[i] = B[lk[now].second];
    }
    REP(i,n) printf("%d ",ans[i]);
    return 0;
}

BZOJ4104 [Thu Summer Camp 2015]解密運算 【亂搞】