UVa 10498 Happiness! （线性规划）

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];

}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