题目描述:
Dr.Kong has entered a bobsled competition because he hopes his hefty weight will give his an advantage over the L meter course (2 <= L<= 1000). Dr.Kong will push off the starting line at 1 meter per second, but his speed can change while he rides along the course. Near the middle of every meter Bessie travels, he can change his speed either by using gravity to accelerate by one meter per second or by braking to stay at the same speed or decrease his speed by one meter per second.
Naturally, Dr.Kong must negotiate N (1 <= N <= 500) turns on the way down the hill. Turn i is located T_i meters from the course start (1 <= T_i <= L-1), and he must be enter the corner meter at a peed of at most S_i meters per second (1 <= S_i <= 1000). Dr.Kong can cross the finish line at any speed he likes.
Help Dr.Kong learn the fastest speed he can attain without exceeding the speed limits on the turns.
Consider this course with the meter markers as integers and the turn speed limits in brackets (e.g., '[3]'):
0 1 2 3 4 5 6 7[3] 8 9 10 11[1] 12 13[8] 14
(Start) |------------------------------------------------------------------------| (Finish)
Below is a chart of Dr.Kong 's speeds at the beginning of each meter length of the course:
Max: [3] [1] [8]
Mtrs: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Spd: 1 2 3 4 5 5 4 3 4 3 2 1 2 3 4
His maximum speed was 5 near the beginning of meter 4.
输入描述:
There are multi test cases,your program should be terminated by EOF
Line 1: Two space-separated integers: L and N
Lines 2..N+1: Line i+1 describes turn i with two space-separated integers: T_i and S_i输出描述:
Line 1: A single integer, representing the maximum speed which Dr.Kong can attain between the start and the finish line, inclusive.样例输入:
14 3
7 3
11 1
13 8样例输出:
5题意分析：
一个长为 l 的路，有n个弯道，在每个弯道的速度不能超过v,
每个时间段可以选择加速或减速或速度不变，求能到达的最大速度。
解题思路：
可以选择一直加速，到达弯道后，如果速度低于要求，可以通过。
如果高于要求，把当前位置速度变成要求的速度，再把前一位置改变成减速状态，直到有足够的距离可以减速。
需要注意给出的弯道位置并不是按照升序排列。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 1020
int c[N];
struct data{
int x;
int y;
}a[N];
int cmp(data a,data b){
return a.x < b.x;
}
int main()
{
int n,l,i,j,s,ans;
while(scanf("%d%d" , &l, &n)!=EOF){
for( i=1; i<=n; i++ )
scanf("%d%d", &a[i].x, &a[i].y);
s=1;
i=1;
sort(a , a+n , cmp);
memset(c, 0 , sizeof(c));
c[0] = 1;
while(s <= l){
c[s] = c[s-1] + 1;
if(s == a[i].x){
if(c[s] > a[i].y){
j = s-1;
c[s] = a[i].y;
while(1){
if(c[j] == c[j+1] + 1 || c[j] == c[j+1])
break;
c[j] = c[j+1] + 1;
j--;
}
}
i++;
}
s++;
}
ans = 0;
for(i=0; i<=l; i++)
if(c[i] > ans)
ans=c[i];
printf("%d
", ans);
}
return 0;
}