CodeForces-545C：Woodcutters（贪心）

最新推荐文章于 2022-05-31 01:51:47 发布

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面对一排树木，每个树都有特定的高度和坐标，木材采伐者的目标是最大化倒下的树的数量，但不能让倒下的树相互重叠。此问题通过一次遍历数组并应用贪心策略解决，最终确定最多能有多少棵树被成功砍伐。

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题目描述

Little Susie listens to fairy tales before bed every day. Today’s fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.

There are n trees located along the road at points with coordinates x1,x2,…,xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi-hi,xi] or [xi;xi+hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn’t contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.

输入

The first line contains integer n (1≤n≤105) — the number of trees.

Next n lines contain pairs of integers xi,hi (1≤xi,hi≤109) — the coordinate and the height of the і-th tree.

The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.

输出

Print a single number — the maximum number of trees that you can cut down by the given rules.

样例输入

5
1 2
2 1
5 10
10 9
19 1

样例输出

提示

In the first sample you can fell the trees like that:
fell the 1-st tree to the left — now it occupies segment [-1;1]
fell the 2-nd tree to the right — now it occupies segment [2;3]
leave the 3-rd tree — it occupies point 5
leave the 4-th tree — it occupies point 10
fell the 5-th tree to the right — now it occupies segment [19;20]

思路：
用贪心的想法，遍历数组，如果一个位置的树可以倒向一边，则实现这种想法，因为本题没有权值，所以不需要进行排序

代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <cctype>
using namespace std;
typedef long long ll;
const int N=1e5+5;
const int Max=0x7f7f7f7f;//必须使用最大值定义
struct tr
{
    int x;
    int h;
}tree[N];

int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
       scanf("%d%d",&tree[i].x,&tree[i].h);
    }
    tree[0].x=-Max;
    tree[n+1].x=Max;
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        if(tree[i].x-tree[i].h>tree[i-1].x)
        {
            sum++;
            continue;
        }
        if(tree[i].x+tree[i].h<tree[i+1].x)
        {
            sum++;
            tree[i].x+=tree[i].h;
        }
    }
    printf("%d\n",sum);
    return 0;
}