POJ 3273 Monthly Expense(二分)

题目链接:点击打开链接

题目大意是说给n个数,要求分成m组,必须连续的数才能合并成一个组,求满足ans大于等于每一组的和的最小ans(每个组可以只有1个数)

显然二分查找最小的ans

//Must so#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<queue>#include<vector>#include<set>#include<map>#include<cstring>#include<cstdio>#include<cmath>#define mem(a,x) memset(a,x,sizeof(a))#define sqrt(n) sqrt((double)n)#define pow(a,b) pow((double)a,(int)b)#define inf (1<<29)#define NN 100005using namespace std;const double PI = acos(-1.0);typedef long long LL;//一样的思想,暴力的遍历每一个可能的答案,check是否满足,找到满足的最小的,于是二分int n,m;//n个数分m组int a[NN];bool ok(int ans)//判断答案能否分m组{    //........分组连续......贪心,只要连续的加起来没有超过ans就可以分到一个组    int k = 1;//傻了...    for (int i = 0,sun = 0;i < n;i++)    {        if (sun + a[i] <= ans) sun += a[i];        else        {            k++;            if (k > m) return 0;//剪枝            sun = a[i];        }    }    return k <= m;//能分6组的话分7组是绰绰有余滴,小于m组的也可以分成m组}int main(){    while (~scanf("%d%d",&n,&m))    {        int l = 0,r = 0;        for (int i = 0;i < n;i++)        {            scanf("%d",a+i);            r += a[i];            l = max(l,a[i]);        }        int mid = (l+r)>>1;        while (l < r)        {            if (ok(mid))//OK只说明mid满足 但是不一定最小                r = mid - 1;//尝试更小的mid            else l = mid + 1;//不满足就尝试更大的mid            mid = (l+r)>>1;        }        cout<<mid<<endl;    }    return 0;}