N-拼图游戏的可解性

NJXZCOJ 844 集训队日常之BigBaiZhou玩游戏

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 #include <stack>
 9 #include <map>
10 #include <set>
11 #include <cmath>
12 #include <cctype>
13 #include <ctime>
14 #include <bitset>
15 
16 using namespace std;
17 
18 typedef long long ll;
19 typedef pair<int, int> pii;
20 const int maxn = 1e3 + 10;
21 int n, m, a;
22 int G[maxn][maxn], c[maxn * maxn];
23 vector<int> v;
24 
25 int lowbit(int x) { return x & (-x); }
26 void update(int x) {
27     while (x < maxn * maxn) { ++c[x]; x += lowbit(x); }
28 }
29 int query(int x) {
30     int ret = 0;
31     while (x > 0) { ret += c[x]; x -= lowbit(x); }
32     return ret;
33 }
34 
35 int main() {
36 #ifdef __AiR_H
37     freopen("A.in", "r", stdin);
38 //    freopen("out.txt", "w", stdout);
39 #endif // __AiR_H
40     while (scanf("%d %d", &n, &m) != EOF) {
41         if (n == 0 && m == 0) { break; } v.clear(); int t;
42         for (int i = 1; i <= n; ++i) {
43             for (int j = 1; j <= m; ++j) {
44                 scanf("%d", &a);
45                 if (a == 0) { t = (n - i) + (m - j); t %= 2; v.push_back(n * m); }
46                 else { v.push_back(a); }
47             }
48         }
49         int Size = v.size(); memset(c, 0, sizeof(c)); ll sum = 0;
50         for (int i = 0; i < Size; ++i) {
51             sum += i - query(v[i]); update(v[i]);
52         }
53 //        printf("%lld\n", sum);
54         if (sum % 2 == t) { printf("YES\n"); } else { printf("NO\n"); }
55     }
56     return 0;
57 }
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