bzoj 3265: 志愿者招募加强版 （单纯形）

3265: 志愿者招募加强版

Time Limit: 20 Sec   Memory Limit: 512 MB
Submit: 537   Solved: 287
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Description

Input

Output

Sample Input

3 3
2 3 4
1 1 2 2
1 2 3 5
1 3 3 2

Sample Output

14

HINT

Source

单纯形

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题解：单纯形裸题啦

#include
#include
#include
#include
#include
#define N 1003
#define M 10003
#define eps 1e-7
#define inf 1000000000
using namespace std;
double c[N],b[M],a[M][N],ans,v;
int n,m;
void priov(int l,int e)
{
	b[l]/=a[l][e];
	for (int i=1;i<=m;i++) 
	 if (i!=e) a[l][i]/=a[l][e];
	a[l][e]=1/a[l][e];
	for (int i=1;i<=n;i++) 
	 if (i!=l&&fabs(a[i][e])>eps) {
	 	b[i]-=b[l]*a[i][e];
	 	for (int j=1;j<=m;j++)
	 	 if (j!=e) a[i][j]-=a[i][e]*a[l][j];
	 	a[i][e]=-a[i][e]*a[l][e];
	 }
	v+=c[e]*b[l];
	for (int i=1;i<=m;i++)
	 if (i!=e) c[i]-=c[e]*a[l][i];
	c[e]=-c[e]*a[l][e];
}
double simple()
{
	int l,i,e;
	double t;
	while (1){
		for (i=1;i<=m;i++) 
		 if (c[i]>eps) break;
		e=i;
		if (e==m+1) return v;
		t=inf;
		for (int i=1;i<=n;i++)
		 if (a[i][e]>eps&&t>b[i]/a[i][e]) 
		  t=b[i]/a[i][e],l=i;
		if (t==inf) return inf;
		priov(l,e);
	}
}
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d%d",&m,&n);
	for (int i=1;i<=m;i++) scanf("%lf",&c[i]);
    for (int i=1;i<=n;i++) {
    	int k; scanf("%d",&k);
    	for (int t=1;t<=k;t++) {
    		int l,r; scanf("%d%d",&l,&r);
    		for (int j=l;j<=r;j++) a[i][j]=1;
		}
		scanf("%lf",&b[i]);
	}
	ans=simple();
	printf("%.0lf\n",ans);
}