UVA - 10498 -线性规划-单纯形法

题目链接：https://vjudge.net/contest/342824#problem/G
题目大意：

// UVa10498 Happiness!
// Rujia Liu
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cassert>
using namespace std;

// 改进单纯性法的实现
// 参考：http://en.wikipedia.org/wiki/Simplex_algorithm
// 输入矩阵a描述线性规划的标准形式。a为m+1行n+1列，其中行0~m-1为不等式，行m为目标函数（最大化）。列0~n-1为变量0~n-1的系数，列n为常数项
// 第i个约束为a[i][0]*x[0] + a[i][1]*x[1] + ... <= a[i][n]
// 目标为max(a[m][0]*x[0] + a[m][1]*x[1] + ... + a[m][n-1]*x[n-1] - a[m][n])
// 注意：变量均有非负约束x[i] >= 0
const int maxm = 500; // 约束数目上限
const int maxn = 500; // 变量数目上限
const double INF = 1e100;
const double eps = 1e-10;

struct Simplex {
  int n; // 变量个数
  int m; // 约束个数
  double a[maxm][maxn]; // 输入矩阵
  int B[maxm], N[maxn]; // 算法辅助变量

  void pivot(int r, int c) {
    swap(N[c], B[r]);
    a[r][c] = 1 / a[r][c];
    for(int j = 0; j <= n; j++) if(j != c) a[r][j] *= a[r][c];
    for(int i = 0; i <= m; i++) if(i != r) {
      for(int j = 0; j <= n; j++) if(j != c) a[i][j] -= a[i][c] * a[r][j];
      a[i][c] = -a[i][c] * a[r][c];
    }
  }

  bool feasible() {
    for(;;) {
      int r, c;
      double p = INF;
      for(int i = 0; i < m; i++) if(a[i][n] < p) p = a[r = i][n];
      if(p > -eps) return true;
      p = 0;
      for(int i = 0; i < n; i++) if(a[r][i] < p) p = a[r][c = i];
      if(p > -eps) return false;
      p = a[r][n] / a[r][c];
      for(int i = r+1; i < m; i++) if(a[i][c] > eps) {
        double v = a[i][n] / a[i][c];
        if(v < p) { r = i; p = v; }
      }
      pivot(r, c);
    }
  }

  // 解有界返回1，无解返回0，无界返回-1。b[i]为x[i]的值，ret为目标函数的值
  int simplex(int n, int m, double x[maxn], double& ret) {
    this->n = n;
    this->m = m;
    for(int i = 0; i < n; i++) N[i] = i;
    for(int i = 0; i < m; i++) B[i] = n+i;
    if(!feasible()) return 0;
    for(;;) {
      int r, c;
      double p = 0;
      for(int i = 0; i < n; i++) if(a[m][i] > p) p = a[m][c = i];
      if(p < eps) {
        for(int i = 0; i < n; i++) if(N[i] < n) x[N[i]] = 0;
        for(int i = 0; i < m; i++) if(B[i] < n) x[B[i]] = a[i][n];
        ret = -a[m][n];
        return 1;
      }
      p = INF;
      for(int i = 0; i < m; i++) if(a[i][c] > eps) {
        double v = a[i][n] / a[i][c];
        if(v < p) { r = i; p = v; }
      }
      if(p == INF) return -1;
      pivot(r, c);
    }
  }
};

//////////////// 题目相关
#include<cmath>
Simplex solver;

int main() {
  int n, m;
  while(scanf("%d%d", &n, &m) == 2) {
    for(int i = 0; i < n; i++) scanf("%lf", &solver.a[m][i]); // 目标函数
    solver.a[m][n] = 0; // 目标函数常数项
    for(int i = 0; i < m; i++)
      for(int j = 0; j < n+1; j++)
        scanf("%lf", &solver.a[i][j]);
    double ans, x[maxn];
    assert(solver.simplex(n, m, x, ans) == 1);
    ans *= m;
    printf("Nasa can spend %d taka.\n", (int)floor(ans + 1 - eps));
  }
  return 0;
}