八数码问题判断是否有解

这道题搜索不会写= =。结果这个是个著名的八数码问题，有一个定理能够判断是否有解：将整个矩阵从上到下从左到右变成一个序列，把0去掉后求出该序列的逆序对x0。若另一序列的逆序对x1满足x0≡x1(mod2)，则有解。

所以就当是复习归并排序求逆序对吧。又写错了的说= =

参考代码

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <iostream>#include <algorithm>#include <vector>#include <string>#include <stack>#include <queue>#include <deque>#include <map>#include <set>using std::cin;using std::cout;using std::endl;inline int readIn(){    int a;    scanf("%d", &a);    return a;}const int maxn = 10005;const char* szAns[] ={    "You still have a chance.",    "You are destined to be single."};int n;int sequeue[maxn];int temp[maxn];int pair;void mergeSort(int l = 0, int r = n){    if (r - l == 1) return;    int mid = (l + r) / 2;    mergeSort(l, mid);    mergeSort(mid, r);    int i = l;    int j = mid;    int k = l;    while (i < mid || j < r)    {        if (j >= r || i < mid && sequeue[i] <= sequeue[j])        {            temp[k++] = sequeue[i++];        }        else        {            temp[k++] = sequeue[j++];            pair += mid - i;        }    }    for (int i = l; i < r; i++)    {        sequeue[i] = temp[i];    }}void run(){    int a = readIn();    while (a--)    {        n = readIn();        int to = n * n;        int cnt = 0;        pair = 0;        for (int i = 1; i <= to; i++)        {            sequeue[cnt] = readIn();            if (sequeue[cnt]) cnt++;        }        mergeSort(0, cnt);        printf("%s\n", szAns[pair & 1]);    }}int main(){    run();    return 0;}

突然看到了牛人的总结，记录一个。