CF1691B Shoe Shuffling

洛谷题面

题目大意

求出每一个大小相同的“块”，如果有块的大小为 \(1\) 则无解。

否则，我们对于每个“块”，设组成此“块”的每个元素的下标为 \([i,i+1,\cdots,i+k]\)，则答案排列为 \([i+k,i,i+1,\cdots,i+k-1]\)。

注意求无解和输出要分开操作。

代码

#include <iostream>
#include <cstdio>
#include <climits>//need "INT_MAX","INT_MIN"
#include <cstring>//need "memset"
#include <numeric>
#include <algorithm>
#include <cmath>
#include <map>
#define enter putchar(10)
#define debug(c,que) std::cerr << #c << " = " << c << que
#define cek(c) puts(c)
#define blow(arr,st,ed,w) for(register int i = (st);i <= (ed); ++ i) std::cout << arr[i] << w;
#define speed_up() std::ios::sync_with_stdio(false),std::cin.tie(0),std::cout.tie(0)
#define mst(a,k) memset(a,k,sizeof(a))
#define stop return(0)
const int mod = 1e9 + 7;
inline int MOD(int x) {
	if(x < 0) x += mod;
	return x % mod;
}
namespace Newstd {
	inline int read() {
		int ret = 0,f = 0;char ch = getchar();
		while (!isdigit(ch)) {
			if(ch == '-') f = 1;
			ch = getchar();
		}
		while (isdigit(ch)) {
			ret = (ret << 3) + (ret << 1) + ch - 48;
			ch = getchar();
		}
		return f ? -ret : ret;
	}
	inline double double_read() {
		long long ret = 0,w = 1,aft = 0,dot = 0,num = 0;
		char ch = getchar();
		while (!isdigit(ch)) {
			if (ch == '-') w = -1;
			ch = getchar();
		}
		while (isdigit(ch) || ch == '.') {
			if (ch == '.') {
				dot = 1;
			} else if (dot == 0) {
				ret = (ret << 3) + (ret << 1) + ch - 48;
			} else {
				aft = (aft << 3) + (aft << 1) + ch - '0';
				num ++;
			}
			ch = getchar();
		}
		return (pow(0.1,num) * aft + ret) * w;
	}
	inline void write(int x) {
		if(x < 0) {
			putchar('-');
			x = -x;
		}
		if(x > 9) write(x / 10);
		putchar(x % 10 + '0');
	}
}
using namespace Newstd;
 
const int N = 1e5 + 5;
int a[N];
int n;
inline void solve() {
	mst(a,0);
	n = read();
	bool mark = true;
	for (register int i = 1;i <= n; ++ i) {
		a[i] = read();
	}
	int L = 1,R = 1;
	while (R <= n) {
		while (R + 1 <= n && a[L] == a[R + 1]) R ++;
		if (L == R) {
			puts("-1");
			return;
		}
		L = R + 1;
		R ++;
	}
	L = R = 1;
	while (R <= n) {
		while (R + 1 <= n && a[L] == a[R + 1]) R ++;
		if (L == R) {
			puts("-1");
			return;
		}
		printf("%d ",R);
		for (register int i = L;i < R; ++ i) printf("%d ",i);
		L = R + 1;
		R ++;
	}
	puts("");
}
int main(void) {
	int T = read();
	while (T --) {
		solve();
	}
	
	return 0;
}