GCJ 2009 Round2 A Crazy Rows

/*  这题之前一度会错意了，没理解“只允许交换相邻的两行”，意味着只能整行交换  以及，最后要保证，每行最后出现1的位置，不得大于该行的行数    又及，这题一开始...没看输入数据，以为输入数据之间是有空格的，相当于一个一个数字输入，后来看完数据格式才发现，其实应该输入一串字符串，然后分离为一个个数字*/

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int MAX_N = 45;int N;int M[MAX_N][MAX_N]; // 矩阵int a[MAX_N]; // a[i]表示第i行最后出现1的位置: 1~n-1 void solve(){fill (a, a + N, -1);int res = 0;//cout << "call solve and res = " << res << endl;for (int i = 0; i < N; i++){a[i] = -1; // 如果第i行不含1的话，就当作-1 for (int j = 0; j < N; j++)if (M[i][j] == 1) a[i] = j;}for (int i = 0; i < N; i++){int pos = -1; // 要移动到第i行的行for (int j = i; j < N; j++){if (a[j] <= i){pos = j;break;} }for (int j = pos; j > i; j--){//cout << "nwo j = " << j << " and pos = " << pos << endl;swap(a[j], a[j - 1]);res++;//cout << "now res = " << res << endl;} }cout << res << endl;} int main(){freopen("E:\\A-large.txt", "r", stdin);freopen("E:\\output2.txt", "w", stdout);int k;cin >> k;for (int kase = 1; kase <= k; kase++ ) {cin >> N;for (int i = 0; i < N; i++){char s[MAX_N];cin >> s;for (int j = 0; j < N; j++)M[i][j] = s[j] - '0';}//for (int i = 0; i < N; i++)//for (int j = 0; j < N; j++)//cin >> M[i][j];printf("Case #%d: ",kase);  solve();}fclose(stdin);fclose(stdout);return 0;}