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.
", ceil(m*ans));
}
return 0;
}
[/cpp]