hdu 4004 二分+ 贪心 The Frog's Games

来源：互联网发布：什么翻译软件好编辑：程序博客网时间：2024/05/21 15:42

题意：

给一列石头，让你找跳m次所需具备的最小能力，即找那么多种m次跳法中最大距离最小的。

解：

二分找答案，哎，做的时候提醒了才想到。决定去做做几道二分的题目。

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;int l,m,n,dis[500010],ant[500010];bool go(int dist){    int pre = ant[1]; int num = 1;    if(pre > dist) return false;    for(int i = 2; i <= n + 1; i ++){        if(pre + ant[i] > dist){pre = ant[i]; num ++;}        else pre += ant[i];        if(num > m) return false;    }    return true;}int main(){    while(~scanf("%d%d%d",&l,&n,&m)){        for(int i = 1; i <= n; i ++)            scanf("%d",dis + i);        sort(dis + 1, dis + 1 + n);        dis[n + 1] = l; dis[0] = 0;        int Max = 0;        for(int i = 1; i <= n + 1; i ++){            ant[i] = dis[i] - dis[i - 1];            Max = max(ant[i],Max);        }//        cout << Max << endl;        int low = Max, high = l;        while(low <= high){            int mid = (low + high) >> 1;            if(go(mid)) high = mid - 1;            else low = mid + 1;        }        printf("%d\n",low);    }    return 0;}