【UOJ#179】线性规划 单纯形

题目链接：

　　http://uoj.ac/problem/179

Solution

就是单纯形模板题，这篇博客就是存一下板子。

Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
using namespace std;
#define eps 1e-9
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
	while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}

int N,M,id[11000];
double a[1100][1100],c[1100],b[1100],v,ans[1100];
inline void Pivot(int l,int e)
{
	swap(id[e],id[N+l]);
	
	double r=a[l][e]; 
	
	a[l][e]=1;
	for (int i=1; i<=N; i++) a[l][i]/=r;
	b[l]/=r;
	
	for (int i=1; i<=M; i++) 
		if (i!=l) {
			r=a[i][e];
			 
			a[i][e]=0,b[i]-=r*b[l];
			for (int j=1; j<=N; j++) a[i][j]-=r*a[l][j];
		}
	
	r=c[e];
	c[e]=0;
	for (int i=1; i<=N; i++) c[i]-=r*a[l][i];
	
	v+=r*b[l];
}

inline int Simplex()
{
	int l,e; double k;
	while (1) {
		l=e=0; k=-eps;
		for (int i=1; i<=M; i++)
			if (b[i]<k) k=b[i],l=i;
		if (!l) break;
		k=-eps;
		for (int i=1; i<=N; i++)
			if (a[l][i]<k && (!e || (rand()&1))) k=a[l][i],e=i;
		if (!e) {puts("Infeasible"); return 1;}
		Pivot(l,e);
	}
	while (1) {
		for (int i=1; i<=N; i++) {
			if (c[i]>eps) {e=i; break;}
			if (i==N) {printf("%.8lf
",v); return 0;}
		}
		double re=1e18; l=0;
		
		for (int i=1; i<=M; i++)
			if (a[i][e]>eps && b[i]/a[i][e]<re)
				re=b[i]/a[i][e],l=i;
				
		if (!l) {puts("Unbounded"); return 2;}
		Pivot(l,e);
	}
	
}

int main()
{
	N=read(),M=read(); int type=read();
	
	for (int i=1; i<=N; i++) scanf("%lf",&c[i]),id[i]=i;
	for (int i=1; i<=M; i++) {
		for (int j=1; j<=N; j++) scanf("%lf",&a[i][j]);
		scanf("%lf",&b[i]);
	}
	
	int flag=Simplex();
	
	if (flag || !type) return 0;
	
	for (int i=1; i<=M; i++) ans[id[N+i]]=b[i];
	for (int i=1; i<=N; i++) printf("%.8lf ",ans[i]);
	
	return 0;
}