CodeForces 551C（集训比赛2B_C题）解题报告

Posted 2020-10-22 caomingpei

题目链接：http://codeforces.com/problemset/problem/551/C

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题意：有n个地点，每个地点有箱子，然后有一定人数，每个人从第i个位置移动到第i+1的位置花费1S，每个人移动一个箱子花费1S，要求把所有箱子全部移除花费的最小时间是多少

思路：二分+贪心的思想，每次对一名学生进行判断。对所求时间进行二分，先让学生到达最远的地方，然后剩下的时间就是用来搬箱子的时间，当学生都用完后，观察是否还有箱子没被搬。通过这种情况调整上下界。

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;int m;
const int MAXN =1e5+10;
ll a[MAXN];
int far =0;
ll sum;

int jud(ll mid){
    int man =0;
    ll s=0;
    for(int i=1;i<=far;i++){
        s+=a[i];
        while(s+i>=mid){
            s-=mid-i;
            man++;
        }
    }
    if(s>0) man++;
    //printf("man:%d\n",man);
    if(man>m)    return 1;
    else    return 0;
    
}

ll proce(){
    ll left =far+1;
    ll right =far+sum;
    ll mid =(left+right)/2;
    while(right>=left){
        mid =(left+right)/2;
        //printf("left:%d right:%d mid:%d\n",left,right,mid);
        if(jud(mid)){
            left =mid+1;
        }else{
            right =mid-1;            
        }
    }
    return left;
}

int main(void){
    scanf("%d %d",&n,&m);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
        if(a[i]!=0)    far =i;
        sum+=a[i];
    }
    ll ans = proce();
    printf("%lld\n",ans);
    return 0;

}

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