拼图游戏的可解性

N-拼图游戏的可解性

---

NJXZCOJ 844 集训队日常之BigBaiZhou玩游戏

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <vector>
#include <stack>
#include <map>
#include <set>
#include <cmath>
#include <cctype>
#include <ctime>
#include <bitset>

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
const int maxn = 1e3 + 10;
int n, m, a;
int G[maxn][maxn], c[maxn * maxn];
vector<int> v;

int lowbit(int x) { return x & (-x); }
void update(int x) {
    while (x < maxn * maxn) { ++c[x]; x += lowbit(x); }
}
int query(int x) {
    int ret = 0;
    while (x > 0) { ret += c[x]; x -= lowbit(x); }
    return ret;
}

int main() {
#ifdef __AiR_H
    freopen("A.in", "r", stdin);
//    freopen("out.txt", "w", stdout);
#endif // __AiR_H
    while (scanf("%d %d", &n, &m) != EOF) {
        if (n == 0 && m == 0) { break; } v.clear(); int t;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                scanf("%d", &a);
                if (a == 0) { t = (n - i) + (m - j); t %= 2; v.push_back(n * m); }
                else { v.push_back(a); }
            }
        }
        int Size = v.size(); memset(c, 0, sizeof(c)); ll sum = 0;
        for (int i = 0; i < Size; ++i) {
            sum += i - query(v[i]); update(v[i]);
        }
//        printf("%lld
", sum);
        if (sum % 2 == t) { printf("YES
"); } else { printf("NO
"); }
    }
    return 0;
}