【BZOJ】【P3265】【志願者招募加強版】【題解】【單純形法】
阿新 • • 發佈:2019-02-08
讀了一天算導,成功yy出單純形~~
Code:
#include<bits/stdc++.h> using namespace std; const int maxn=10010; const int maxm=1010; const double eps=1e-8; double a[maxn][maxm]; int n,m; int dcmp(double x){ if(fabs(x)<eps)return 0; return x>0?1:-1; } double simplex(bool f); void pivot(int l,int e,bool f){ for(int i=0;i<=m+f;i++){ if(i==e)continue; a[l][i]/=a[l][e]; }a[l][e]=1/a[l][e]; for(int i=0;i<=n+f;i++){ if(i==l)continue; for(int j=0;j<=m+f;j++){ if(j==e)continue; a[i][j]-=a[i][e]*a[l][j]; }a[i][e]*=-a[l][e]; } } void init(){ int pos=-1; double minn=1e10; for(int i=1;i<=n;i++) if(dcmp(minn-a[i][0])==1)minn=a[i][0],pos=i; if(dcmp(minn)>=0)return; for(int i=0;i<=m;i++)swap(a[0][i],a[n+1][i]); for(int i=0;i<=n;i++)a[i][m+1]=-1; pivot(pos,m+1,1); if(!dcmp(simplex(1))){ for(int i=1;i<=m+1;i++)if(!dcmp(a[0][i]+1)){pos=i;break;} for(int i=0;i<=m+1;i++)swap(a[0][i],a[n+1][i]); for(int i=0;i<=n+1;i++)swap(a[i][pos],a[i][m+1]); }else puts("Can't solve!!"); } double simplex(bool f){ for(;;){ int pos=-1,l,e; double minn=0; for(int i=1;i<=m+f;i++) if(dcmp(a[0][i]-minn)==1)minn=a[0][i],pos=i; if(!~pos)return -a[0][0]; e=pos;pos=-1;minn=1e10; for(int i=1;i<=n;i++) if(dcmp(a[i][e])==1&&dcmp(a[i][0]/a[i][e]-minn)==-1) minn=a[i][0]/a[i][e],pos=i; if(!~pos){ printf("INF!!\n"); return 1e10; }l=pos; pivot(l,e,f); } } int main(){ scanf("%d%d",&m,&n); for(int i=1;i<=m;i++)scanf("%lf",&a[0][i]); for(int i=1;i<=n;i++){ int k;scanf("%d",&k); while(k--){ int l,r;scanf("%d%d",&l,&r); for(int j=l;j<=r;j++)a[i][j]=1; }scanf("%d",&k); a[i][0]=k; } init(); int ans=simplex(0)+.5; printf("%d\n",ans); return 0; }