POJ 1661 Help Jimmy

        这算是一个经典的动态规划题了吧,由于是中文题目,所以题目很好理解.

 

        这个题目的关键也就是找出状态转移方程,然后跟据它写出递归函数(当然也可以用迭代),我越来越意识到递归是个很神奇的东西,和分治法联系在一起的时候问题一下子就变得豁然开朗了.

 

        由于下落的时间是一定的,所以只需要求出最短的爬行时间再加上下落时间就可以了.

        大概思路就是下面的:

 

                               0,   k=N+1

FindMin(x,k) =         max{x  -  aPlatform[k].Lx  +  FindMin(  aPlatform[k].Lx  ,  NextLeft  ),

                                       aPlatform[k].Rx  -  x  +  FindMin(  aPlatform[k].Rx  ,  NextRight  )     },  0<=k<=N

 

 

#include<stdio.h>#include<string.h>#include<stdlib.h>#define MAX 1000#define INFINITE 10000000int t,n,x,y,max,N;struct platform{ int Lx,Rx,h;};struct platform aPlatform[MAX + 5];int aLeftMinTime[MAX + 5];int aRightMinTime[MAX + 5];int FindBoard(int x,int k){ int i; for( i = k + 1 ; i <= N + 1 ; i++ ) { if( aPlatform[i].Lx <= x && aPlatform[i].Rx >= x ) return i; }}int MyCompare(const void *e1,const void *e2){ struct platform *p1,*p2; p1 = (struct platform *)e1; p2 = (struct platform *)e2; return p2->h - p1->h;}int FindMin(int x,int k){ int NextLeft,NextRight; int Lsum,Rsum; if( k == N + 1 ) return 0; NextLeft = FindBoard( aPlatform[k].Lx , k ); NextRight = FindBoard( aPlatform[k].Rx , k ); if( aPlatform[k].h - aPlatform[NextLeft].h > max ) aLeftMinTime[k] = INFINITE; if( aPlatform[k].h - aPlatform[NextRight].h > max ) aRightMinTime[k] = INFINITE; if( aLeftMinTime[k] == -1 ) aLeftMinTime[k] = FindMin(aPlatform[k].Lx,NextLeft); if( aRightMinTime[k] == -1 ) aRightMinTime[k] = FindMin(aPlatform[k].Rx,NextRight); Lsum = x - aPlatform[k].Lx + aLeftMinTime[k]; Rsum = aPlatform[k].Rx - x + aRightMinTime[k]; if( Lsum < Rsum ) return Lsum; else return Rsum;}int main(void){ int i; while(scanf("%d",&t) != EOF) { while(t--) { memset( aLeftMinTime , -1 , sizeof(aLeftMinTime) ); memset( aRightMinTime , -1 , sizeof(aRightMinTime) ); scanf("%d%d%d%d",&N,&x,&y,&max); aPlatform[0].Lx = x; aPlatform[0].Rx = x; aPlatform[0].h = y; aPlatform[N+1].Lx = -20001; aPlatform[N+1].Rx = 20001; aPlatform[N+1].h = 0; for( i = 1 ; i <= N ; i++ ) scanf("%d%d%d",&aPlatform[i].Lx,&aPlatform[i].Rx,&aPlatform[i].h); qsort( aPlatform , N + 1 ,sizeof(struct platform) ,MyCompare ); printf("%d/n",FindMin(aPlatform[0].Lx,0)+y ); } } return 0;}

 

 

 

 

 

还可以参照这个的解释,算是很详尽的了:

http://www.cppblog.com/mythit/archive/2009/06/26/88539.aspx