绝境求生(八数码有解无解的问题,求逆序)

问题 E: 【排序】绝境求生

时间限制: 1 Sec 内存限制: 64 MB
提交: 19 解决: 12

题目描述

The Eight Puzzle, among other sliding-tile puzzles, is one of the famous problems in artificial intelligence. Along with chess, tic-tac-toe and backgammon, it has been used to study search algorithms.

The Eight Puzzle can be generalized into an M × N Puzzle where at least one of M and N is odd. The puzzle is constructed with MN − 1 sliding tiles with each a number from 1 to MN − 1 on it packed into a M by N frame with one tile missing. For example, with M = 4 and N = 3, a puzzle may look like:

Let's call missing tile 0. The only legal operation is to exchange 0 and the tile with which it shares an edge. The goal of the puzzle is to find a sequence of legal operations that makes it look like:

The following steps solve the puzzle given above.

Given an M × N puzzle, you are to determine whether it can be solved.

样例输入

复制样例数据

样例输出

YES
NO

结论:N*M矩阵中，若M为奇数->计算给出矩阵的逆序数和与初始状态的逆序数看奇偶性:奇偶相同则存在，反之不存在

若M为偶数->计算给出矩阵的逆序数为ans1，初始0所在行与给出的0所在行之间的行间距ans2，判断:

（ans1%2） ==（ans2%2）,成立则存在，反之不存在

用树状数组维护

/**/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <stack>
#include <queue>typedef long long LL;
using namespace std;int n, m;
int cnt;
int c[1000005], a[1000005];
int ans;int lowbit(int x){return x & (-x);
}void add(int x){while(x <= cnt){c[x]++;x += lowbit(x);}
}int sum(int x){int su = 0;while(x){su += c[x];x -= lowbit(x);}return su;
}int main()
{//freopen("in.txt", "r", stdin);//freopen("out.txt", "w", stdout);while(scanf("%d %d", &n, &m) == 2){if(!n && !m) break;memset(c, 0, sizeof(c));int p, x, ans = 0;cnt = 0;for (int i = 1; i <= n; i++){for (int j = 1; j <= m; j++){scanf("%d", &x);if(x == 0) p = n - i;else{a[cnt++] = x;}}}for (int i = cnt - 1; i >= 0; i--){ans += sum(a[i]);//cout << a[i] << " " <<sum(a[i]) << endl;add(a[i]);}if(m & 1){if(ans % 2 == 0){printf("YES\n");}else{printf("NO\n");}continue;}if(ans % 2 == p % 2){printf("YES\n");}else{printf("NO\n");}}return 0;
}
/**/