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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define N 1003
#define M 10003
#define eps 1e-7
#define inf 1000000000
using namespace std;
double c
,b[M],a[M]
,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);
}