UVaLive 7455 Linear Ecosystem (Gaussi 消元)

题意：对一个k元向量, 每次左乘一个k*k的矩阵得到新的向量.问经过一定次数的左乘后，能否使得该向量不再变化. (同时要求此时向量非零)。

析：设初始向量为A，矩阵为P.由于每次矩阵P都是左乘A, 那么可以把若干个P合并. 则题目的条件是：

化简为：由于要求所以 P-1 必须不可逆.可以直接用高斯消元求P-1的秩，判断是否可逆（满秩即可逆）.

所以这个题，并不用求解，只要判断秩就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e2 + 5;

const int mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

double a[25][25];

bool Gauss(){

    int ans = 0, r = 0;

    for(int i = 0; i < n; ++i){

        for(int j = r; j < n; ++j)

            if(fabs(a[j][i]) > eps){

                for(int k = i; k < n; ++k)

                    swap(a[j][k], a[r][k]);

                break;

            }

        if(fabs(a[r][i]) < eps){  ++ans;  continue; }

        for(int j = 0; j < n; ++j)

            if(j != r && fabs(a[j][i]) > eps){

                double tmp = a[j][i]/a[r][i];

                for(int k = i; k < n; ++k)

                    a[j][k] -= tmp * a[r][k];

            }

        ++r;

    }

    return ans;

}

int main(){

    int T;  cin >> T;

    for(int kase = 1; kase <= T; ++kase){

        scanf("%d", &n);

        for(int i = 0; i < n; ++i){

            for(int j = 0; j < n; ++j)

                scanf("%lf", a[i]+j);

            a[i][i] -= 1.0;

        }

        printf("%d", Gauss());

        if(kase == T) continue;

        printf("%c", kase % 5 ? ' ' : '\n');

    }

    if(T % 5)  printf("\n");

    return 0;

}

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UVaLive 7455 Linear Ecosystem (Gaussi 消元)

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