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  • 【BZOJ】1689: [Usaco2005 Open] Muddy roads 泥泞的路(贪心)

    http://www.lydsy.com/JudgeOnline/problem.php?id=1689

    一开始我也想到了贪心,,,策略是如果两个连续的水池的距离小于l的话,那么就将他们链接起来,,,然后全部操作完后直接在每个大联通块上除以长度然后取木板就行了。。

    但是不知道哪里错了T_T

    正解:放木板尽量往后放。证明:假设当前i有水池,i-1没有水池,因为长度是固定的,那么从i放显然由于从i-1放木板。

    然后模拟放木板即可。

    优化:算出每个连续的水池需要放的数量,然后加上即可

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <string>
    #include <iostream>
    #include <algorithm>
    #include <queue>
    using namespace std;
    #define rep(i, n) for(int i=0; i<(n); ++i)
    #define for1(i,a,n) for(int i=(a);i<=(n);++i)
    #define for2(i,a,n) for(int i=(a);i<(n);++i)
    #define for3(i,a,n) for(int i=(a);i>=(n);--i)
    #define for4(i,a,n) for(int i=(a);i>(n);--i)
    #define CC(i,a) memset(i,a,sizeof(i))
    #define read(a) a=getint()
    #define print(a) printf("%d", a)
    #define dbg(x) cout << #x << " = " << x << endl
    #define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
    inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
    inline const int max(const int &a, const int &b) { return a>b?a:b; }
    inline const int min(const int &a, const int &b) { return a<b?a:b; }
    
    const int N=10005;
    struct dat { int x, y; }a[N];
    bool cmp(const dat &a, const dat &b) { return a.x<b.x; }
    int n, l;
    
    int main() {
    	read(n); read(l); 
    	for1(i, 1, n) read(a[i].x), read(a[i].y);
    	sort(a+1, a+1+n, cmp);
    	int ans=0, now=0;
    	for1(i, 1, n) {
    		if(now>=a[i].y) continue;
    		now=max(now, a[i].x);
    		int t=a[i].y-now, t2=t/l;
    		if(t%l==0) ans+=t2, now+=t2*l;
    		else ans+=t2+1, now+=(t2+1)*l;
    	}
    	print(ans);
    	return 0;
    }
    

    Description

    Farmer John has a problem: the dirt road from his farm to town has suffered in the recent rainstorms and now contains (1 <= N <= 10,000) mud pools. Farmer John has a collection of wooden planks of length L that he can use to bridge these mud pools. He can overlap planks and the ends do not need to be anchored on the ground. However, he must cover each pool completely. Given the mud pools, help FJ figure out the minimum number of planks he needs in order to completely cover all the mud pools.

        牧场里下了一场暴雨,泥泞道路上出现了许多水坑,约翰想用一批长度 为L的木板将这些水坑盖住.    牧场里的道路可以看成一根数轴,每个水坑可以用数轴上的两个坐标表示,如(3,6)表示从3到6有一个长度为3的水 坑.所有的水坑都是不重叠的,(3,6)和(6,9)可以出现在同一个输入数据中,因为它们是两个连续的水坑,但不重叠.
        请你帮助约翰计算最少要用多少块木板才能将所有水坑盖住

    Input

    * Line 1: Two space-separated integers: N and L * Lines 2..N+1: Line i+1 contains two space-separated integers: s_i and e_i (0 <= s_i < e_i <= 1,000,000,000) that specify the start and end points of a mud pool along the road. The mud pools will not overlap. These numbers specify points, so a mud pool from 35 to 39 can be covered by a single board of length 4. Mud pools at (3,6) and (6,9) are not considered to overlap.

        第1行有二个用空格隔开的整数N和L.其中1≤N≤10000,表示水坑总数.L为木板长度.
    接下来的N行每行有二个用整数si和ei(0≤si<ei≤109),表示一个水坑的两个坐标.

    Output

    * Line 1: The miminum number of planks FJ needs to use.

     
        一个整数,表示约翰盖住所有水坑最少要用多少块长为L的木板.

    Sample Input

    3 3
    1 6
    13 17
    8 12

    Sample Output

    5

    HINT


    Source

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  • 原文地址:https://www.cnblogs.com/iwtwiioi/p/3963125.html
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