《动态规划》递推，递归，HelpJimmy（知道动态规划是什么，解决那种问题，有些逻辑不太好想）

https://www.jianshu.com/p/7799f6ede3f2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

/*
记忆递归的程序
*/

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAX_N 1000
#define INFINITE 1000000
int t, n, x, y, maxHeight;
struct Platform
{
    int Lx, Rx, h;
    bool operator<(const Platform &p2) const
    {
        return h > p2.h;
    }
};

Platform platForms[MAX_N + 10];
int leftMinTime[MAX_N + 10];
int rightMinTime[MAX_N + 10];
int L[MAX_N + 10];

int MinTime(int I, bool bLeft)
{
    int y = platForms[I].h;
    int x;
    if (bLeft)
        x = platForms[I].Lx;
    else
        x = platForms[I].Rx;
    int i;
    for (i = I + 1; i <= n; i++)
    {
        // make sure the downstairs is effective
        if (platForms[i].Lx <= x && platForms[i].Rx >= x)
            break;
    }
    if (i <= n)
    {
        if (y - platForms[i].h > maxHeight)
            return INFINITE;
    }
    else// why else is here?
    {
        if (y > maxHeight)
            return INFINITE;
        else
            return y;

    }
    int nLeftTime = y - platForms[i].h + x - platForms[i].Lx;
    int nRightTime = y - platForms[i].h + platForms[i].Rx -x;
    if (leftMinTime[i] == -1)
        leftMinTime[i] = MinTime(i, true);
    if (L[i] == -1)
    {
        L[i] = MinTime(i, false);
    }
    nLeftTime += leftMinTime[i];
    nRightTime += L[i];
    if (nLeftTime < nRightTime)
        return nLeftTime;
    return nRightTime;
}

int main()
{
    scanf_s("%d", &t);
    for (int i = 0; i < t; i++)
    {
        memset(leftMinTime, -1, sizeof(leftMinTime));
        memset(L, -1, sizeof(rightMinTime));
        scanf_s("%d%d%d%d", &n, &x, &y, &maxHeight);
        platForms[0].Lx = x;
        platForms[0].Rx = x;
        platForms[0].h = y;
        for (int j = 1; j <= n; j++)
            scanf_s("%d%d%d", &platForms[j].Lx, &platForms[j].Rx, &platForms[j].h);
        sort(platForms, platForms + n + 1);
        printf("%d\n", MinTime(0, true));
    }
    return 0;
}