B. Batch Sort

http://codeforces.com/contest/724/problem/B

被坑了,一开始以为如果有一行已经是排好序了,然后有一行需要转换的次数 >= 2的话,那就直接no了。

因为一开始以后转换次数>=2必定需要用了转换列,然后就搞乱了排序好的哪一行,但是排序好的那一行可以用交换两个数来修复。

2 4

1  2 3 4

2 1  4 3

然后其他的思路就是,①、有一个需要转换3次以上的就不行了。

②、随便找一个转换次数为2的,开始暴力枚举他们两两用列转换,然后再判断是否全部只需要一次转换就行了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;

#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
const int maxn = 20 + 20;
int a[maxn][maxn];
int n, m;
int mx;
bool check() {
    bool flag = false;
    for (int i = 1; i <= n; ++i) {
        int j;
        for (j = 1; j <= m; ++j) {
            if (a[i][j] != j) {
                break;
            }
        }
        if (j == m + 1) {
            flag = true;
            break;
        }
    }
    if (flag == false) return false;
    for (int i = 1; i <= n; ++i) {
        int t = 0;
        for (int j = 1; j <= m; ++j) {
            if (a[i][j] != j) ++t;
        }
//        if (t >= 3) return true;
        mx = max(t, mx);
    }
    if (mx >= 5) return true;
    return false;
}
int row[maxn];
void getrow() {
    for (int i = 1; i <= n; ++i) {
        row[i] = 0;
        for (int j = 1; j <= m; ++j) {
            row[i] += a[i][j] != j;
        }
    }
}
bool allone() {
    for (int i = 1; i <= n; ++i) {
        if (row[i] >= 3) return false;
    }
    return true;
}
int gg[maxn];
void get(int aa, int bb) {
    for (int i = 1; i <= n; ++i) {
        swap(a[i][aa], a[i][bb]);
    }
}
void show() {
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            cout << a[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
}
void work() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            cin >> a[i][j];
        }
    }

    if (check() || mx >= 5) {
        cout << "NO" << endl;
        return;
    }
    getrow();
    if (allone()) {
        cout << "YES" << endl;
        return;
    }
    int pos;
//    show();
    for (int i = 1; i <= n; ++i) {
        if (row[i] >= 3) {
            pos = i;
            break;
        }
    }
    int lengg = 0;
    for (int i = 1; i <= m; ++i) {
        if (a[pos][i] != i) {
            gg[++lengg] = i;
        }
    }
//    for (int i = 1; i <= m; ++i) {
//        cout << gg[i] << " ";
//    }
//    cout << endl;
    for (int i = 1; i <= lengg; ++i) {
        for (int j = i + 1; j <= lengg; ++j) {
            get(gg[i], gg[j]);
            getrow();
            if (allone()) {
                cout << "YES" << endl;
                return;
            }
            get(gg[i], gg[j]);
        }
    }
    cout << "NO" << endl;
    return;
}
int main() {
#ifdef local
    freopen("data.txt","r",stdin);
#endif
    work();
    return 0;
}
View Code
原文地址:https://www.cnblogs.com/liuweimingcprogram/p/5940681.html