NYOJ 309 BOBSLEDDING（dp）
BOBSLEDDING
時間限制：1000?ms ?|? 內(nèi)存限制：65535?KB 難度：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.
輸入
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
題意：一個人在滑雪的過程中可以每秒加速1m/s，也可以保持速度不變，還可以每秒減速1m/s。滑道長L米，有n個拐角，當(dāng)人滑行到拐角i時，他的速度不能超過Si。問在整個滑行過程中，人的最大速度是多少。
分析：假設(shè)人可以一直加速，從前往后遍歷每個區(qū)間，當(dāng)?shù)侥硞€拐角時的速度大于這個拐角的最大速度時，就向前更新前面的點(diǎn)的最大速度。dp[i]表示在第i米時可以滑行的最大速度。
#include<stdio.h> #include<string.h> int main() {int L, n, i, dp[1005];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 pos = 1, lim = 1;for(i = 0; i <= L; i++){if(dp[i] == 0) //不是拐角的地方一直加速dp[i] = lim;else{if(dp[i] > dp[i-1])dp[i] = dp[i-1] + 1;else{pos = i; //從當(dāng)前位置向前更新while(dp[pos - 1] - dp[pos] > 1){dp[pos-1] = dp[pos] + 1;pos--;}}}lim = dp[i] + 1;}int mmax = 0;for(i = 0; i <= L; i++)if(dp[i] > mmax)mmax = dp[i];printf("%d\n",mmax);}return 0; }
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