[bzoj3112][zjoi2013]防守防线

3112: [Zjoi2013]防守战线

Time Limit: 20 Sec Memory Limit: 512 MB
Submit: 1420 Solved: 822
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sol:
论单纯形的优越性

#include<cstdio>#include<algorithm>#include<string>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>using namespace std;int n,m;const double eps=1e-8;inline int read(){    char c;    int res,flag=0;    while((c=getchar())>'9'||c<'0') if(c=='-')flag=1;    res=c-'0';    while((c=getchar())>='0'&&c<='9') res=(res<<3)+(res<<1)+c-'0';    return flag?-res:res;}int l,e;double a[1100][11000];double c[1100],d[11000];double v=0;inline void povit(){    c[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(fabs(a[i][e])>eps&&i!=l)    {        for(int j=1;j<=m;++j) if(j!=e) a[i][j]-=a[l][j]*a[i][e];        c[i]-=c[l]*a[i][e];        a[i][e]*=-a[l][e];    }    v+=d[e]*c[l];    for(int i=1;i<=m;++i)    if(i!=e) d[i]-=d[e]*a[l][i];    d[e]*=-a[l][e];}inline double simplex(){    while(true)    {        l=0;e=0;        for(e=1;e<=m;++e) if(d[e]>eps) break;        if(e==m+1) return v;        double save=1e9;        for(int i=1;i<=n;++i)        if(a[i][e]>eps&&save>c[i]/a[i][e]) save=c[i]/a[i][e],l=i;         if(save==1e9) return 1e9;        povit();    }}int main(){//  freopen("3112.in","r",stdin);//  freopen(".out","w",stdout);    n=read();    m=read();    for(int i=1;i<=n;++i) scanf("%lf",&c[i]);    int l,r;    for(int i=1;i<=m;++i)    {        l=read();        r=read();        for(int j=l;j<=r;++j) a[j][i]=1;        scanf("%lf",&d[i]);    }    printf("%d",(int)(simplex()+0.5));}