nyoj 309 乘雪橇（模拟）_nyoj 309 bobsledding-优快云博客

本文链接：https://blog.youkuaiyun.com/linyuxilu/article/details/51457024

本文介绍了一种算法解决雪橇竞赛中如何最大化速度同时确保在每个转弯处不超过速度限制的问题。通过更新速度数组来保证在转弯前适当减速，并最终找到整个赛程中的最高速度。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

BOBSLEDDING

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 3

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

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

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

样例输出

分析：

题意就是从起点滑到终点，中间有一些转弯，转弯时有速度限制，而且在滑行过程中没一米只能匀加速或者匀减速，所以当转弯时需要提前开始减速，才能顺利通过，所以当遇到转弯，速度过大时，就需要更新前面的数据，才能符合条件。

主要就是对数组做处理，当不符合条件是更新数据，等到所有数据确定以后再寻找最大值，如果是边更新边处理，得到的可能不是有效的最大值

#include<stdio.h>
#include<string.h>
int main()
{
    int l,n,i,dp[1010];
    while(~scanf("%d%d",&l,&n))
    {
        int a,b;
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&a,&b);
            dp[a]=b;
        }
        int p=1,lim=1;
        for(i=0; i<=l; i++)
        {
            if(dp[i]==0)//不是拐弯的地方一直加速
                dp[i]=lim;
            else
            {
                if(dp[i]>dp[i-1])//拐弯地方
                    dp[i]=dp[i-1]+1;
                else//如果到达拐弯处速度过大，就要更新前面的
                {
                    p=i;//从第i个开始更新
                    while(dp[p-1]-dp[p]>1)
                    {
                        dp[p-1]=dp[p]+1;//一直更新到符合条件
                        p--;
                    }
                }
            }
            lim=dp[i]+1;//速度加1
        }
        int max=0;//寻找最大速度
        for(i=0; i<=l; i++)
            if(dp[i]>max)
                max=dp[i];
        printf("%d\n",max);
    }
    return 0;
}