POJ 1661

#include <iostream>#include <vector>#include <algorithm>#include <cstring>using namespace std;const int MAXLEN = 2000;const int INFINIT = 1000000;struct Platform{int Lx;int Rx;int h;};vector<Platform> platforms;int LeftMinTime[MAXLEN];int RightMinTime[MAXLEN];int n,x,y,imax;//递归求解int MinTime(int L,bool bLeft){int y=platforms[L].h;int x;if(bLeft)x=platforms[L].Lx;elsex=platforms[L].Rx;int i;for(i = L + 1; i <= n; i++)if(x >= platforms[i].Lx && x <= platforms[i].Rx)break;if(i <= n){if(y - platforms[i].h > imax)return INFINIT;}else{if(y > imax)return INFINIT;else return y;}int iLeftTime = y - platforms[i].h + x - platforms[i].Lx;int iRighttime = y - platforms[i].h + platforms[i].Rx - x;if(LeftMinTime[i] == -1)LeftMinTime[i] = MinTime(i, true);if(RightMinTime[i] == -1)RightMinTime[i] = MinTime(i, false);iLeftTime += LeftMinTime[i];iRighttime += RightMinTime[i];if(iLeftTime > iRighttime)return iRighttime;return iLeftTime;} //for cmparebool cmpare(const Platform & v1, const Platform & v2){return v1.h>v2.h;}int main(){int t;while(cin>>t && (t >= 0 && t <= 20)){for(int i = 0; i < t; i++){memset(LeftMinTime, -1, sizeof(LeftMinTime));memset(RightMinTime, -1, sizeof(RightMinTime));cin>>n>>x>>y>>imax;Platform temp;temp.Lx = x;temp.Rx = x;temp.h = y;platforms.push_back(temp);for(int j = 0; j<n; j++){cin>>temp.Lx>>temp.Rx>>temp.h;platforms.push_back(temp);}sort(platforms.begin(),platforms.end(), cmpare);cout<<MinTime(0,true)<<endl;platforms.erase(platforms.begin(),platforms.end());}}return 0;}

解题报告：

1、个人认为动态规划较难题，状态转移不容易找，找着后也不容易写出代码

2、可以这样思考这道题，其子问题是从板的左边到地面的最短时间和从板的右端到地面的最
   短时间，为此可以为板编号，从上到下编号，其中小球的编程为0，这样依次编下去，
   在同一高度编号前后无所谓，于是该题的状态是木板的编号，值为从该板的左端到地面和
   板的右端到地面的最短时间。

3、注意的地方是在用vector时不要忘了在最后用下erase方法，对其进行清空