poj_1681_高斯消元

这道题和之前的把那一道1222很类似。仅仅只是一定要注意一下对于无解的推断。

/*#########################################################################
# File Name: poj_1681.cpp
# Author: CaoLei
# Created Time: 2015/7/20 18:33:08
#########################################################################*/

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <set>
#include <queue>
#include <map>
using namespace std;
#define MAX(x,y) (((x)>(y))?(x):(y))
#define MIN(x,y) (((x)<(y))?(x):(y))
#define N 500010
#define pi acos(-1.0)
#define inf 100000000
typedef long long ll;
typedef unsigned long long ull;
int a[300][300];
char str[20];
int n;
bool f;
void gauss(){
    for (int i = 0; i<n*n; i++){
        int p=i;
        for (int j = i; j<n*n; j++){          //这个地方j的初始值是i不
        //是0，错了一次切记。
            if (a[p][i]<a[j][i]){
                p=j;
            }
        }
        for (int j = 0; j <= n*n; j++) swap(a[i][j], a[p][j]);
        int count=0;
        for(int k=i;k<n*n;k++)       //推断是否有解，关键
          if(a[i][k]) count++;
        if(!count&&a[i][n*n]){
            f=false;
            return ;
        }
        for (int j = 0; j<n*n; j++)
          if (i != j&&a[j][i])
            for (int k = i + 1; k <= n*n; k++)
              a[j][k] ^= a[i][k];

    }
}

int main(){
    freopen("in.txt", "r", stdin);
    int t;
    scanf("%d", &t);
    while (t--){
        f = true;
        memset(a, 0, sizeof(a));
        scanf("%d", &n);
        for (int i = 0; i<n; i++){
            scanf("%s",str);
            for (int j = 0; j<n; j++){
                if (str[j] == 'w'){
                    a[i*n + j][n*n] = 1;
                }
            }
        }
        for (int j = 0; j<n*n; j++){
            a[j][j] = 1;
            if (j%n != 0) a[j - 1][j] = 1;
            if ((j + 1) % n != 0) a[j + 1][j] = 1;
            if (j >= n) a[j - n][j] = 1;
            if (j<(n - 1)*n) a[j + n][j] = 1;
        }
        gauss();
        if (!f){
            printf("inf\n");
            continue;
        }
        int count = 0;
        for (int i = 0; i<n*n; i++){
            if(a[i][n*n]) count++;
        }
        printf("%d\n",count);
    }
    return 0;
}