博弈论（最少交换次数使序列有序）-优快云博客

文章描述了一个涉及二维数组的游戏规则，通过C++代码实现判断是否能赢得游戏并计算操作次数。主要函数包括初始化、状态检查、操作次数计算和最终决策。

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 205;
int a[maxn][maxn], n, m;
int r[maxn], c[maxn], b[maxn], maxr[maxn], maxc[maxn], minr[maxn], minc[maxn];
bool vis[maxn];
void print(int x){
    if(x == -1) cout << "NSFW\n";
    else if(x == 0) cout << "FOX\n";
    else cout << "CAT\n";
}
void init(){
    for(int i = 1; i <= maxn - 5; i++){
        maxr[i] = maxc[i] = 0;
        minr[i] = minc[i] = 1e9;
    }
}
int op_times(int n){
    for(int i = 1; i <= n; i++){
        vis[i] = 0;
    }
    int cnt = 0;
    for(int i = 1; i <= n; i++){
        if(!vis[i]){
            cnt++;
            int j = i;
            while(!vis[j]){
                vis[j] = 1;
                j = r[j];
            }
        }
    }
    return n - cnt;
}
bool no_win(){
    for(int j = 1; j <= m; j++){
        for(int i = 1; i <= n; i++){
            if(i >= 2 && j >= 2)
              if((a[i][j] - a[i][j - 1]) != (a[i - 1][j] - a[i - 1][j - 1])){
                  return 1;
              } 
            maxr[i] = max(maxr[i], a[i][j]);
            maxc[j] = max(maxc[j], a[i][j]);
            minr[i] = min(minr[i], a[i][j]);
        }
    }
    int i;
    for(i = 1; i <= n; i++){
        r[i] = maxr[i];
        b[i] = r[i];
    }
    sort(b + 1, b + n + 1);
    for(int i = 1; i <= n; i++){
        r[i] = lower_bound(b + 1, b + n + 1, r[i]) - b;
    }
    for(int i = 1; i <= n; i++){
        if(minr[i] != (r[i] - 1) * m + 1 || maxr[i] != r[i] * m){//不用minr判断也可以
            return 1;
        } 
    }
    return 0;
}
bool check(){
    int last = 0;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(a[i][j] < last) return 0;
            last = a[i][j];
        }
    }
    return 1;
}
void solve(){
    int i, j;
    cin >> n >> m;
    init();
    for(i = 1; i <= n; i++){
        for(j = 1; j <= m; j++){
            cin >> a[i][j];
        }
    }
    if(no_win()){
        print(-1);
        return;
    }
    int cntr = op_times(n), cntc;
    for(j = 1; j <= m; j++){
        r[j] = maxc[j];
        b[j] = r[j];
    }
    sort(b + 1, b + m + 1);
    for(int i = 1; i <= m; i++){
        r[i] = lower_bound(b + 1, b + m + 1, r[i]) - b;
    }
    cntc = op_times(m);
    if(cntr == 1 && cntc == 0){
        print(0);
        return;
    }
    if(n == 2 && m == 2){
        if((cntr + cntc) % 2 == 0) print(1);
        else print(0);
    }
    else if(n == 2){
        if((cntr + cntc) % 2 == 0) print(1);
        else print(-1);
    }
    else if(m == 2){
        if((cntr + cntc) % 2 == 1) print(0);
        else print(-1);
    }
    else print(-1);
}
int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while(T--){
        solve();
    }
    return 0;
}