POJ 2437

题意:给定n个泥坑路面的初末位置,判断最少需要几个桥才能覆盖这些路面。

贪心法,从每个泥坑的开始位置建桥,然后判断是否覆盖后面的路面,就是分类讨论了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <algorithm>
using namespace std;

struct Seg{
	int l, r;
	Seg(int l = 0, int r = 0) :l(l), r(r){}
	bool operator < (const Seg & rhs) const {
		return l < rhs.l;
	}
};
Seg v[10240];

int main()
{
    ios::sync_with_stdio(false);
	int n, l; cin >> n >> l;
	for (int i = 0; i < n; ++i) cin >> v[i].l >> v[i].r;
	sort(v, v + n);
	int cnt = 0, pos = 0;
	for (int i = 0; i < n; ++i){
		if (pos >= v[i].r) continue;
		if (pos <= v[i].l) {
			int len = v[i].r - v[i].l;
			int cur = (len + l - 1) / l;
			pos = v[i].l + cur * l;
			cnt += cur;
		}
		else{
			int len = v[i].r - pos;
			int cur = (len + l - 1) / l;
			pos += cur * l;
			cnt += cur;
		}
	}
	cout << cnt << endl;
	return 0;
}


原文地址:https://www.cnblogs.com/kunsoft/p/5312673.html