POJ 1661 Help Jimmy

题目链接：http://poj.org/problem?id=1661

分析：类似于最长递增子序列的dp题；坑点：如果下落过程中碰到板，那么必须在改板向左右再下落，不可以穿板，所以找到一次，该方向(左或者右）就不需要再找了

#include<iostream>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<numeric>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<cctype>
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int NINF = -INF - 1;
const int maxn = 1e3 + 5;
typedef long long ll;
#define MOD 1000000007
using namespace std;
int dp[maxn][2];
struct node
{
    int x1, x2, h;
    friend bool operator < (const node &m, const node &n)
    {
        return m.h < n.h;
    }
}a[maxn];
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int n, x, y, maxm;
        scanf("%d %d %d %d", &n, &x, &y, &maxm);
        for (int i = 0; i <= n + 1; ++i) dp[i][0] = dp[i][1] = INF;
        for (int i = 1; i <= n; ++i)
        {
            int p, q, c;
            scanf("%d %d %d", &p, &q, &c);
            a[i] = node{p, q, c};
        }
        a[0] = node{-20000, 20000, 0};
        a[n + 1] = node{x, x, y};
        sort(a, a + n + 1);
//        for (int i = 0; i <= n + 1; ++i) cout << a[i].x1 << ' ' << a[i].x2 << ' ' << a[i].h << endl;
        for (int i = 1; i <= n + 1; ++i)
        {
            int flag1 = 0, flag2 = 0;
            for (int j = i - 1; j >= 0; --j)
            {
                if (a[i].h - a[j].h > maxm) break;
                if (flag1 && flag2) break;
                if (!flag1)
                {
                    if (j == 0)
                    {
                        dp[i][0] = a[i].h;
                        flag1++;
                    }
                    else {
                        if (a[j].x1 <= a[i].x1 && a[j].x2 >= a[i].x1)
                        {
                            dp[i][0] = a[i].h - a[j].h + min(dp[j][0] + a[i].x1 - a[j].x1, dp[j][1] + a[j].x2 - a[i].x1);
                            flag1++;
                        }
                    }
                }
                if (!flag2)
                {
                    if (j == 0)
                    {
                        dp[i][1] = a[i].h;
                        flag2++;
                    }
                    else {
                        if (a[j].x1 <= a[i].x2 && a[j].x2 >= a[i].x2)
                        {
                            dp[i][1] = a[i].h - a[j].h + min(dp[j][0] + a[i].x2 - a[j].x1, dp[j][1] + a[j].x2 - a[i].x2);
                            flag2++;
                        }
                    }
                }
            }
//            cout << i << ' ' << dp[i][0] << ' ' << dp[i][1] << endl;
        }
//        cout << dp[n + 1][0] << ' ' << dp[n + 1][1];
        printf("%d
", min(dp[n + 1][0], dp[n + 1][1]));
    }
    return 0;
}