题意

将N种食品分给m个参赛选手，一个单位的某食品给某个选手一定满足度，每个选手有一个最大满足度。为了避免浪费，分给每一个选手的食品都不超越选手的满足度。已知的各种食品的单价，求最多可以花的钱。

思路

标准的线性规划---直接求即可……目前只会用用模板，还没搞清楚单纯性算法……

代码

  [cpp] /** Simplex C(n+m)(n) maximize: c[1]*x[1]+c[2]*x[2]+...+c[n]*x[n]+ans subject to a[1,1]*x[1]+a[1,2]*x[2]+...a[1,n]*x[n] <= rhs[1] a[2,1]*x[1]+a[2,2]*x[2]+...a[2,n]*x[n] <= rhs[2] ...... a[m,1]*x[1]+a[m,2]*x[2]+...a[m,n]*x[n] <= rhs[m] 限制: 传入的矩阵必须是标准形式的, 即目标函数要最大化；约束不等式均为<= ；xi为非负数(>=0). simplex返回参数： OPTIMAL 有唯一最优解 UNBOUNDED 最优值无边界 FEASIBLE 有可行解 INFEASIBLE 无解 n为元素个数,m为约束个数 线性规划： max c[]*x; a[][]<=rhs[]; ans即为结果,x[]为一组解(最优解or可行解) **/ const double eps = 1e-8; const double inf = 1e15; #define OPTIMAL -1 //表示有唯一的最优基本可行解 #define UNBOUNDED -2 //表示目标函数的最大值无边界 #define FEASIBLE -3 //表示有可行解 #define INFEASIBLE -4 //表示无解 #define PIVOT_OK 1 //还可以松弛 #define maxn 1000 struct LinearProgramming{ int basic[maxn], row[maxn], col[maxn]; double c0[maxn]; double dcmp(double x){ if (x > eps) return 1; else if (x < -eps) return -1; return 0; } void init(int n, int m, double c[], double a[maxn][maxn], double rhs[], double &ans) { //初始化 for(int i = 0; i <= n+m; i++) { for(int j = 0; j <= n+m; j++) a[i][j]=0; basic[i]=0; row[i]=0; col[i]=0; c[i]=0; rhs[i]=0; } ans=0; } //转轴操作 int Pivot(int n, int m, double c[], double a[maxn][maxn], double rhs[], int &i, int &j){ double min = inf; int k = -1; for (j = 0; j <= n; j ++) if (!basic[j] && dcmp(c[j]) > 0) if (k < 0 || dcmp(c[j] - c[k]) > 0) k = j; j = k; if (k < 0) return OPTIMAL; for (k = -1, i = 1; i <= m; i ++) if (dcmp(a[i][j]) > 0) if (dcmp(rhs[i] / a[i][j] - min) < 0){ min = rhs[i]/a[i][j]; k = i; } i = k; if (k < 0) return UNBOUNDED; else return PIVOT_OK; } int PhaseII(int n, int m, double c[], double a[maxn][maxn], double rhs[], double &ans, int PivotIndex){ int i, j, k, l; double tmp; while(k = Pivot(n, m, c, a, rhs, i, j), k == PIVOT_OK || PivotIndex){ if (PivotIndex){ i = PivotIndex; j = PivotIndex = 0; } basic[row[i]] = 0; col[row[i]] = 0; basic[j] = 1; col[j] = i; row[i] = j; tmp = a[i][j]; for (k = 0; k <= n; k ++) a[i][k] /= tmp; rhs[i] /= tmp; for (k = 1; k <= m; k ++) if (k != i && dcmp(a[k][j])){ tmp = -a[k][j]; for (l = 0; l <= n; l ++) a[k][l] += tmp*a[i][l]; rhs[k] += tmp*rhs[i]; } tmp = -c[j]; for (l = 0; l <= n; l ++) c[l] += a[i][l]*tmp; ans -= tmp * rhs[i]; } return k; } int PhaseI(int n, int m, double c[], double a[maxn][maxn], double rhs[], double &ans){ int i, j, k = -1; double tmp, min = 0, ans0 = 0; for (i = 1; i <= m; i ++) if (dcmp(rhs[i]-min) < 0){min = rhs[i]; k = i;} if (k < 0) return FEASIBLE; for (i = 1; i <= m; i ++) a[i][0] = -1; for (j = 1; j <= n; j ++) c0[j] = 0; c0[0] = -1; PhaseII(n, m, c0, a, rhs, ans0, k); if (dcmp(ans0) < 0) return INFEASIBLE; for (i = 1; i <= m; i ++) a[i][0] = 0; for (j = 1; j <= n; j ++) if (dcmp(c[j]) && basic[j]){ tmp = c[j]; ans += rhs[col[j]] * tmp; for (i = 0; i <= n; i ++) c[i] -= tmp*a[col[j]][i]; } return FEASIBLE; } //standard form //n:原变量个数 m:原约束条件个数 //c:目标函数系数向量-[1~n],c[0] = 0; //a:约束条件系数矩阵-[1~m][1~n] rhs:约束条件不等式右边常数列向量-[1~m] //ans:最优值 x:最优解||可行解向量-[1~n] int simplex(int n, int m, double c[], double a[maxn][maxn], double rhs[], double &ans, double x[]){ int i, j, k; //标准形式变松弛形式 for (i = 1; i <= m; i ++){ for (j = n+1; j <= n+m; j ++) a[i][j] = 0; a[i][n+i] = 1; a[i][0] = 0; row[i] = n+i; col[n+i] = i; } k = PhaseI(n+m, m, c, a, rhs, ans); if (k == INFEASIBLE) return k; k = PhaseII(n+m, m, c, a, rhs, ans, 0); for (j = 0; j <= n+m; j ++) x[j] = 0; for (i = 1; i <= m; i ++) x[row[i]] = rhs[i]; return k; } }ps; //Primal Simplex int n,m; double c[maxn], ans, a[maxn][maxn], b[maxn], x[maxn]; int main(){ //freopen("test.in", "r", stdin); //freopen("test.out", "w", stdout); while(scanf("%d %d", &n, &m) != EOF){ double ans; ps.init(n, m, c, a, b, ans); for (int i = 1; i <= n; i ++) scanf("%lf", &c[i]); for (int i = 1; i <= m; i ++){ for (int j = 1; j <= n; j ++){ scanf("%lf", &a[i][j]); } scanf("%lf", &b[i]); } ps.simplex(n,m,c,a,b,ans,x); printf("Nasa can spend %.0f taka.\n", ceil(m*ans)); } return 0; } [/cpp]
举杯独醉，饮罢飞雪，茫然又一年岁。 ------AbandonZHANG