HDU 3359 高斯消元模板题，

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HDU 3359 高斯消元模板题，

2024-09-10 00:33:55 220人阅读

http://acm.hdu.edu.cn/showproblem.php?pid=3359

题目的意思是，由矩阵A生成矩阵B的方法是:

以a[i][j]为中心的，哈曼顿距离不大于dis的数字的总和 / 个数，就是矩阵B的b[i][j]

现在给出B，要求A

那么我们设A矩阵为a[1][1], a[1][2], a[1][3].....

那么对于每一个b[i][j]我们有b[i][j] = (a[1][1] + a[1][2] + ... + ) / cnt

所以这样可以建议一条方程，然后guass求解。

注意题目的输出格式，printf("%8.2lf")后，不需要加空格。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 2e2 + 20;
const double eps = 1e-7;
class GaussMatrix {
public:
    double a[maxn][maxn];
    int equ, val; //方程个数（行），和变量个数（列），其中第val个是b值，不能取
    void swapRow(int rowOne, int rowTwo) {
        for (int i = 1; i <= val; ++i) {
            swap(a[rowOne][i], a[rowTwo][i]);
        }
    }
    void swapCol(int colOne, int colTwo) {
        for (int i = 1; i <= equ; ++i) {
            swap(a[i][colOne], a[i][colTwo]);
        }
    }
    bool same(double x, double y) {
        return fabs(x - y) < eps;
    }
    int guass() {
        int k, col;
        for (k = 1, col = 1; k <= equ && col < val; ++k, ++col) { //不能取到第val个
            int maxRow = k; //选出列最大值，误差最小
            for (int i = k + 1; i <= equ; ++i) {
                if (fabs(a[i][col]) > fabs(a[maxRow][col])) {
                    maxRow = i;
                }
            }
            if (same(a[maxRow][col], 0)) {
                --k;
                continue;
            }
            if (maxRow != k) swapRow(k, maxRow);
            for (int i = col + 1; i <= val; ++i) { //约去系数
                a[k][i] /= a[k][col];
            }
            a[k][col] = 1.0; //第一个就要变成1了，然后下面和上面的变成0
            for (int i = 1; i <= equ; ++i) {
                if (i == k) continue; //当前这行，不操作
                for (int j = col + 1; j <= val; ++j) {
                    a[i][j] -= a[i][col] * a[k][j];
                }
                a[i][col] = 0.0;
            }
//            debug();
        }
        for (k; k <= equ; ++k) {
            if (!same(a[k][val], 0)) return -1; //方程无解
        }
        return val - k; //自由变量个数
    }
    void debug() {
        for (int i = 1; i <= equ; ++i) {
            for (int j = 1; j <= val; ++j) {
                printf("%6.2lf ", a[i][j]);
            }
            printf("\n");
        }
        printf("*******************************************\n\n");
    }
}arr;
int dis;
double mp[maxn][maxn];
int vis[maxn][maxn], DFN;
int n, m;
int tonext[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
struct bfsNode {
    int cnt, x, y;
    bfsNode(int _cnt, int _x, int _y) {
        cnt = _cnt, x = _x, y = _y;
    }
};
queue<struct bfsNode>que;
int toHash(int x, int y) {
    return x * max(n, m) + y;
}
void init(int row, int col, int which) {
    ++DFN;
    while (!que.empty()) que.pop();
    arr.a[which][toHash(row, col)] = 1.0;
    que.push(bfsNode(0, row, col));
    vis[row][col] = DFN;
    int has = 1;
    while (!que.empty()) {
        struct bfsNode t = que.front();
        que.pop();
        if (t.cnt + 1 > dis) break;
        for (int i = 0; i < 4; ++i) {
            int tx = t.x + tonext[i][0], ty = t.y + tonext[i][1];
            if (tx >= 1 && tx <= n && ty >= 1 && ty <= m && vis[tx][ty] != DFN) {
                vis[tx][ty] = DFN;
                arr.a[which][toHash(tx, ty)] = 1.0;
                que.push(bfsNode(t.cnt + 1, tx, ty));
                has++;
            }
        }
    }
    arr.a[which][toHash(n, m) + 1] = mp[row][col] * has;
}
void work() {
    n = arr.equ, m = arr.val;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            scanf("%lf", &mp[i][j]);
        }
    }
    int which = 0;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            init(i, j, ++which);
        }
    }
    arr.equ = which;
    arr.val = toHash(n, m) + 1;
//    arr.debug();
    arr.guass();
//    arr.debug();
    int to = 1;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            printf("%8.2lf", arr.a[to++][toHash(n, m) + 1]);
        }
        printf("\n");
    }
}
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    while (scanf("%d%d%d", &arr.val, &arr.equ, &dis) != EOF && arr.val + arr.equ + dis) {
        if (!flag) printf("\n");
        flag = false;
        work();
        memset(&arr, 0, sizeof arr);
    }
    return 0;
}

View Code

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <assert.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;


#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <bitset>
const int maxn = 1e2 + 20;
const double eps = 1e-7;
class GaussMatrix {
public:
    double a[maxn][maxn], x[maxn];
    int equ, val; //方程个数（行），和变量个数（列），其中第val个是b值，不能取
    void swapRow(int rowOne, int rowTwo) {
        for (int i = 1; i <= val; ++i) {
            swap(a[rowOne][i], a[rowTwo][i]);
        }
    }
    void swapCol(int colOne, int colTwo) {
        for (int i = 1; i <= equ; ++i) {
            swap(a[i][colOne], a[i][colTwo]);
        }
    }
    bool same(double x, double y) {
        return fabs(x - y) < eps;
    }
    int guass() {
        int k, col;
        for (k = 1, col = 1; k <= equ && col < val; ++k, ++col) { //不能取到第val个
            int maxRow = k; //选出列最大值，误差最小
            for (int i = k + 1; i <= equ; ++i) {
                if (fabs(a[i][col]) > fabs(a[maxRow][col])) {
                    maxRow = i;
                }
            }
            if (same(a[maxRow][col], 0)) {
                --k;
                continue;
            }
            if (maxRow != k) swapRow(k, maxRow);
            for (int i = col + 1; i <= val; ++i) { //约去系数
                a[k][i] /= a[k][col];
            }
            a[k][col] = 1.0; //第一个就要变成1了，然后下面和上面的变成0
            for (int i = 1; i <= equ; ++i) {
                if (i == k) continue; //当前这行，不操作
                for (int j = col + 1; j <= val; ++j) {
                    a[i][j] -= a[i][col] * a[k][j];
                }
                a[i][col] = 0.0;
            }
            debug();
        }
        for (k; k <= equ; ++k) {
            if (!same(a[k][val], 0)) return -1; //方程无解
        }
        return val - k; //自由变量个数
    }
    void debug() {
        for (int i = 1; i <= equ; ++i) {
            for (int j = 1; j <= val; ++j) {
                printf("%6.2lf ", a[i][j]);
            }
            printf("\n");
        }
        printf("*******************************************\n\n");
    }
}arr;
void work() {
//    arr.equ = 3, arr.val = 5;
//    arr.a[1][1] = 1, arr.a[1][2] = 2, arr.a[1][3] = -1, arr.a[1][4] = 1, arr.a[1][5] = 2;
//    arr.a[2][1] = 2, arr.a[2][2] = -1, arr.a[2][3] = 1, arr.a[2][4] = -3, arr.a[2][5] = -1;
//    arr.a[3][1] = 4, arr.a[3][2] = 3, arr.a[3][3] = -1, arr.a[3][4] = -1, arr.a[3][5] = 3;
//    int res = arr.guass();
//    cout << res << endl;

//    arr.equ = 4, arr.val = 3 + 1;
//    arr.a[1][1] = 2, arr.a[1][2] = 3, arr.a[1][3] = 1, arr.a[1][4] = 4;
//    arr.a[2][1] = 1, arr.a[2][2] = -2, arr.a[2][3] = 4, arr.a[2][4] = -5;
//    arr.a[3][1] = 3, arr.a[3][2] = 8, arr.a[3][3] = -2, arr.a[3][4] = 13;
//    arr.a[4][1] = 4, arr.a[4][2] = -1, arr.a[4][3] = 9, arr.a[4][4] = -6;
//    cout << arr.guass() << endl;

//    arr.equ = 3, arr.val = 3 + 1;
//    arr.a[1][1] = 2, arr.a[1][2] = 3, arr.a[1][3] = 1, arr.a[1][4] = 16;
//    arr.a[2][1] = 1, arr.a[2][2] = 5, arr.a[2][3] = 2, arr.a[2][4] = 23;
//    arr.a[3][1] = 3, arr.a[3][2] = 4, arr.a[3][3] = 5, arr.a[3][4] = 33;
//    cout << arr.guass() << endl;

//    arr.equ = 3, arr.val = 4;
//    arr.a[1][1] = 2, arr.a[1][2] = 3, arr.a[1][3] = -1, arr.a[1][4] = 2;
//    arr.a[2][1] = 3, arr.a[2][2] = -2, arr.a[2][3] = 1, arr.a[2][4] = 2;
//    arr.a[3][1] = 1, arr.a[3][2] = -5, arr.a[3][3] = 2, arr.a[3][4] = 1;
//    cout << arr.guass() << endl;

    arr.equ = 3, arr.val = 4;
    arr.a[1][1] = 1, arr.a[1][2] = 2, arr.a[1][3] = 3, arr.a[1][4] = 4;
    arr.a[2][1] = 1, arr.a[2][2] = 0, arr.a[2][3] = 2, arr.a[2][4] = 5;
    arr.a[3][1] = 0, arr.a[3][2] = 0, arr.a[3][3] = 5, arr.a[3][4] = 5;
    cout << arr.guass() << endl;
}
int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    work();
    return 0;
}

高斯消元模板 && 题目

HDU 3359 高斯消元模板题，

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HDU 3359 高斯消元模板题，

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