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.
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.
Output
* Line 1: The miminum number of planks FJ needs to use.
Sample Input
1 6
13 17
8 12
Sample Output
HINT
sb模拟题
用pair<int,int>写的……感觉骚的没朋友
#include<cstdio> #include<iostream> #include<cstring> #include<cstdlib> #include<algorithm> #include<cmath> #include<queue> #include<deque> #include<set> #include<map> #include<ctime> #define LL long long #define inf 0x7ffffff #define pa pair<int,int> #define pi 3.1415926535897932384626433832795028841971 using namespace std; inline LL read() { LL x=0,f=1; char ch= getchar (); while (ch< '0' ||ch> '9' ){ if (ch== '-' )f=-1;ch= getchar ();} while (ch>= '0' &&ch<= '9' ){x=x*10+ch- '0' ;ch= getchar ();} return x*f; } int n,m,l,ans; pa a[10010]; int main() { n=read();m=read(); for ( int i=1;i<=n;i++) { a[i].first=read(); a[i].second=read()-1; } sort(a+1,a+n+1); for ( int i=1;i<=n;i++) { l=max(l,a[i].first-1); int len=a[i].second-a[i].first; ans+=len/m; l+=(len/m)*m; while (l<a[i].second)l+=m,ans++; } printf ( "%d\n" ,ans); } |
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