poj_1681_高斯消元

標籤：數學   高斯消元   

這道題和之前的把那一道1222非常相似，只不過一定要注意一下對於無解的判斷。

/*########################################################################## 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 100000000typedef 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++){          //這個地方j的初始值是i不        //是0，錯了一次切記。            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++)       //判斷是否有解，關鍵          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;}

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poj_1681_高斯消元