ZOJ2002
题目
Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. One of the most famous scribers lived in the 15th century and his name was Xaverius Endricus Remius Ontius Xendrianus (Xerox). Anyway, the work was very annoying and boring. And the only way to speed it up was to hire more scribers.
Once upon a time, there was a theater ensemble that wanted to play famous Antique Tragedies. The scripts of these plays were divided into many books and actors needed more copies of them, of course. So they hired many scribers to make copies of these books. Imagine you have m books (numbered 1, 2 ... m) that may have different number of pages (p1, p2 ... pm) and you want to make one copy of each of them. Your task is to divide these books among k scribes, k <= m. Each book can be assigned to a single scriber only, and every scriber must get a continuous sequence of books. That means, there exists an increasing succession of numbers 0 = b0 < b1 < b2, ... < bk-1 <= bk = m such that i-th scriber gets a sequence of books with numbers between bi-1+1 and bi. The time needed to make a copy of all the books is determined by the scriber who was assigned the most work. Therefore, our goal is to minimize the maximum number of pages assigned to a single scriber. Your task is to find the optimal assignment.
题意
给一段不下降的子序列，要求分成k段，使每一段的和的最大值最小，若有多组答案，输出依次满足：第一组和尽量小，第二组和尽量小，第三组和尽量小。。。的答案。
思路
二分+贪心，二分出来的段数如果小于k，就从前向后分段。
注意二分上下界！！！
代码
#include<cstdio>
#include<cstring>
#include<iostream>
#include<iomanip>
#include<algorithm>
#define ll long long
#include<cmath>
#define exp 0.000000001
using namespace std;
ll T,m,k,cntt;
ll a[520],f[520];
bool check(ll x){
memset(f,false,sizeof(f));
ll cur=0,cnt=1;
for(int i=m;i>=1;i--){
cur+=a[i];
if(cur>x){
cnt++;
cur=a[i];
f[i]=true;
}
}
cntt=cnt;
return cnt<=k;
}
int main(){
cin>>T;
while(T--){
ll l=0,r=0;
scanf("%lld%lld",&m,&k);
for(ll i=1;i<=m;i++){
scanf("%lld",&a[i]);
if(a[i]>l) l=a[i];//此处很关键
r+=a[i];
}
ll mid,ans;
while(l<=r){
mid=(l+r)>>1;
if(check(mid)){
r=mid-1;
ans=mid;
}
else{
l=mid+1;
}
}
check(ans);
//cout<<"!!"<<cntt<<endl;
for(int i=1;i<=m&&cntt<k;i++){
if(!f[i]) f[i]=true,cntt++;
}
for(int i=1;i<=m;i++){
cout<<a[i];
if(i!=m) cout<<' ';
if(f[i]) cout<<"/ ";
}
cout<<endl;
}
return 0;
}