洛谷P2514||bzoj2426 [HAOI2010]工厂选址

洛谷P2514

bzoj2426

其实是个简单的贪心，然而不适合在脑子不清醒的时候做...看不懂题意续了1个小时

很容易发现应该枚举新建哪个发电厂，对于这种方案就是取其中b吨煤运到原来发电厂，取剩下(suma-b)吨煤运到新发电厂。首先假设全部都运到原来发电厂，然后把其中suma-b吨代价最小的改为运到新发电厂即可

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 ll m,b,n;
 ll h[];
 ll a[],c[],d[],e[];
 ll s1,s2,s3,s;
 ll ansx,ansy=0x3f3f3f3f3f3f3f3f;
 bool c1(ll x,ll y)
 {
     return d[x]-c[x]<d[y]-c[y];
 }
 int main()
 {
     ll i,j,t;
     scanf("%lld%lld%lld%lld",&m,&b,&h[],&n);
     for(i=;i<=m;++i)
     {
         scanf("%lld",&a[i]);
         s+=a[i];
     }
     s-=b;
     for(i=;i<=n;++i)
         scanf("%lld",&h[i]);
     for(i=;i<=m;++i)
     {
         scanf("%lld",&c[i]);
         s1+=a[i]*c[i];
     }
     for(i=;i<=n;++i)
     {
         for(j=;j<=m;++j)
             scanf("%lld",&d[j]);
         for(j=;j<=m;++j)
             e[j]=j;
         sort(e+,e+m+,c1);
         s2=s1;s3=;
         for(j=;j<=m;++j)
         {
             t=min(a[e[j]],s-s3);
             s2+=t*(d[e[j]]-c[e[j]]);
             s3+=t;
         }
         s2+=h[];s2+=h[i];
         if(s2<ansy)    {ansy=s2;ansx=i;}
     }
     printf("%lld\n%lld\n",ansx,ansy);
     return ;
 }