编程之美2015资格赛 C.基站选址

这一题看到题目就感觉是三分，然后将基站位于每个格子的总代价打表观察发现，每一行是凹函数or单调减，每一列也是凹函数or单调减，将每一行的最小值取出来组成一行，也是凹函数or单调减。所以用三分嵌套即可。

先对行三分，在计算每一行的最小值时也用三分搜索。

注意这里面可能会爆int，所以要用long long存总代价。

其实我这里面的三分有点问题，对于单调增的数列找到的是第二小的值==，取决于是 if(a[mid]<a[midr]) r=midr;还是if(a[mid]<=a[midr]) r=midr;
不过最后也过了== 改日研究研究一个更好的写法。

#include<iostream>
#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<vector>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<ctype.h>
#include<map>
#include<time.h>
#include<bitset>
#include<set>
#include<list>
using namespace std;
//BOP quali Pro3

//range of number:
//int  2147483647; 10^9
//int   -2147483648;
//long long  9223372036854775807; 10^18
//long long   -9223372036854775808;
//sqrt(INT_MAX)=46340
const int maxn=1010;
int T;
int N;
int M;
int A;
int B;
long long x;
long long y;
pair<long long,long long>user[maxn];
pair<long long,long long>comp[maxn];
long long ans;
long long disuser(long long x1,long long y1,long long x2,long long y2)
{
return (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}
long long discomp(long long x1,long long y1,long long x2,long long y2)
{
return abs(x1-x2)+abs(y1-y2);
}
long long cal(long long i,long long j)
{
long long tmp=0;
for(int k=0;k<A;k++)
{
tmp+=disuser(i,j,user[k].first,user[k].second);
}
long long tmp2=0x3f3f3f3f;
for(int k=0;k<B;k++)
{
tmp2=min(tmp2,discomp(i,j,comp[k].first,comp[k].second));
}
tmp+=tmp2;
return tmp;
}
long long trisearch(int row)
{
//cout<<"r: "<<row<<endl;
int l=1;
int r=M;
while(l<r-1)
{
int mid=(r+l)/2;
int midr=(mid+r)/2;
//cout<<mid<<" "<<midr<<endl;
// cout<<cal(row,mid)<<" "<<cal(row,midr)<<endl;
// system("pause");
if(cal(row,mid)<cal(row,midr))
r=midr;
else
l=mid;
}
//cout<<cal(row,l)<<" "<<row<<" "<<l<<endl;
return min(cal(row,l),cal(row,r));

}
void solve()
{
int l=1;
int r=N;
while(l<r-1)
{
int mid=(r+l)/2;
int midr=(mid+r)/2;
//cout<<mid<<" "<<trisearch(mid)<<" "<<midr<<" "<<trisearch(midr)<<endl;
if(trisearch(mid)<trisearch(midr))
r=midr;
else
l=mid;
//cout<<mid<<" "<<midr<<endl;
}
ans=min(trisearch(l),trisearch(r));
}
void test()
{
for(int i=1;i<=N;i++)
{
//    int aa=0x3f3f3f3f;
for(int j=1;j<=M;j++)
{
//if(mp[i][j]!=0) continue;
long long tmp=0;
for(int k=0;k<A;k++)
{
tmp+=disuser(i,j,user[k].first,user[k].second);
}
long long tmp2=0x3f3f3f3f;
for(int k=0;k<B;k++)
{
tmp2=min(tmp2,discomp(i,j,comp[k].first,comp[k].second));
}
tmp+=tmp2;
ans=min(ans,tmp);
//            aa=min(aa,tmp);
//cout<<tmp<<", ";
}
//cout<<aa<<endl;
// cout<<endl;
}
}

int testtri()
{
int a[26]={10397, 9742, 9135, 8576, 8065, 7604, 7191, 6826, 6509, 6240, 6019, 5846, 5721, 5643, 5612, 5629, 5694, 5807, 5968, 6177, 6436, 6743, 7098, 7501, 7952, 8451};
//int a[13]={99,88,77,66,55,44,33,32,31,10,8,6,1};
//int a[13]={0,1,3,5,6,7,9,11,14,16,17,23,25};
int l=1;
int r=12;
while(l<r-1)
{
int mid=(r+l)/2;
int midr=(mid+r)/2;
if(a[mid]<a[midr])// can not deal with ascending array
{
r=midr;
}
else l=mid;
cout<<mid<<" m "<<midr<<endl;
}
cout<<a[l]<<" "<<a[r]<<endl;
return a[l]<a[r]?a[l]:a[r];
}
int main()
{

scanf("%d",&T);
for(int ca=1;ca<=T;ca++)
{
scanf("%d %d %d %d",&N,&M,&A,&B);
memset(user,0,sizeof(user));
memset(user,0,sizeof(comp));
ans=0x3f3f3f3f;
for(int i=0;i<A;i++)
{
scanf("%lld %lld",&x,&y);
user[i]=make_pair(x,y);
}
for(int i=0;i<B;i++)
{
scanf("%lld %lld",&x,&y);
comp[i]=make_pair(x,y);
}
solve();
printf("Case #%d: %lld\n",ca,ans);
}
return 0;
}

偷偷看了一下kuangbin大神的代码，似乎这样的三分可以同时处理凹函数，单调增/减函数。

//int a[]={1,3,6,9,13,14,46,78,88};
//int a[]={88,78,46,14,13,9,6,3,1};
int a[]={88,78,46,1,3,6,9,13,14};
int l=0;
int r=8;
int x1=0;
while(l <= r)
{
int mid1 = l + (r-l)/3;
int mid2 = r - (r-l)/3;
if(a[mid1]<= a[mid2])
{
x1 = mid1;
r = mid2-1;
}
else{
x1 = mid2;
l = mid1+1;
}
}
int ans=min(a[l],a[r]);
cout<<ans<<endl;
//cout<<a[x1]<<" "<<a[l]<<" "<<a[r]<<endl;