UVa OJ 714 - Copying Books

UVa OJ 714 - Copying Books

Problem

把一個包含m個正整數的序列劃分成k個（1≤k≤m≤500）非空的連續子序列，使得每個正 整數恰好屬於一個序列。設第i個序列的各數之和爲S(i)，你的任務是讓所有S(i)的最大值盡 量小。例如，序列1 2 3 2 5 4劃分成3個序列的最優方案爲1 2 3 | 2 5 | 4，其中S(1)、S(2)、S(3) 分別爲6、7、4，最大值爲7；如果劃分成1 2 | 3 2 | 5 4，則最大值爲9，不如剛纔的好。每個 整數不超過107。如果有多解，S(1)應儘量小。如果仍然有多解，S(2)應儘量小，依此類推

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly two lines. At the first line, there are two integers m and k, 1 ≤ k ≤ m ≤ 500. At the second line, there are integers p1, p2, … , pm separated by spaces. All these values are positive and less than 10000000.

Output

For each case, print exactly one line. The line must contain the input succession p1, p2, … pm divided
into exactly k parts such that the maximum sum of a single part should be as small as possible. Use
the slash character (‘/’) to separate the parts. There must be exactly one space character between any
two successive numbers and between the number and the slash.
  If there is more than one solution, print the one that minimizes the work assigned to the first scriber,
then to the second scriber etc. But each scriber must be assigned at least one book.

Sample Input

2
9 3
100 200 300 400 500 600 700 800 900
5 4
100 100 100 100 100

Sample Output

100 200 300 400 500 / 600 700 / 800 900
100 / 100 / 100 / 100 100

Solution

這道題目基本就是二分加貪心，先確定二分的最值，然後從右往左一步步貪心求範圍。不過要記得題目有限制必須是K個區域。

#include <iostream>
#include <cstring>
using namespace std;

const int maxn = 505;
long long num[maxn], sum[maxn];
char idx[maxn],a[maxn];
long long leftt,rightt,mid;
int m,k,n;

inline bool binSearch(){
    long long val = 0; int cnt = 0;
    memset(idx, 0, sizeof(idx));
    for(int i = m; i >= 1; --i){
        if(val+num[i] <= mid && i >= k-cnt)
            val += num[i];
        else{
            cnt++;
            val = num[i];
            idx[i] = 1;
        }
    }
    if(cnt == k-1){
        memcpy(a, idx ,sizeof(idx));
        return true;
    }
    return false;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int cas;
    cin >> cas;
    while(cas--){
        rightt = leftt = 0;
        cin >> m >> k;
        for(int i = 1; i <= m; ++i){
            cin >> num[i];
            rightt += num[i];
            leftt = num[i]>leftt ? num[i]:leftt;
        }
        while(rightt >= leftt){
            mid = (leftt+rightt) >> 1;
            if(binSearch()) rightt = mid -1;
            else leftt = mid+1;
        }
        for(int i = 1; i < m; ++i){
            cout << num[i] << ' ';
            if (a[i] == 1) cout << '/' << ' ';
        }
        cout << num[m] << '\n';
    }
    return 0;
}