POJ 3273 （二分&三分_E题）解题报告

题目链接：http://poj.org/problem?id=3273

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题意：将一个长为M的数串按序分为k组，使得每组和最小。

思路：二分最小化和的题目，二分的标准是每组的和，判断标准是组数。如果组数大于要求的，证明每组和较小，反之，证明每组和较大，利用二分的方法求得。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
typedef long long ll;
const double PI = acos(-1.0);
const double eps = 1e-6;
int N =0;
int M =0;
const int MAXN =1e5+10;
int a[MAXN];
int high=0;
int low =0;

int jud(int d){
    int sum =0;
    int count =0;
    for(int i=0;i<N;i++){
        if(sum+a[i]<=d){
            sum+=a[i];
        }else if(a[i]<=d){
            count++;
            sum =a[i];
        }else if(a[i]>d){return 1;}
    }

    if(count>=M) return 1;
    else return 0;
}

int bsearch(){
    int left = low;
    int right =high;
    int mid;
    while(right>=left){
        mid = (right-left)*0.5+left;
        if(jud(mid)){
            left=mid+1;
        }else{
            right =mid-1;
        }
    }
    return left;
}

int main(void){

    scanf("%d %d",&N,&M);

    for(int i=0;i<N;i++){
        scanf("%d",&a[i]);
        high+=a[i];
        low=min(low,a[i]); 
    }

    printf("%d\n",bsearch());
    return 0;

}