POJ 1661 Help Jimmy

Posted 2022-11-18 veasky

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

 

 1 #include<iostream>
 2 #include<sstream>
 3 #include<cstdio>
 4 #include<cstdlib>
 5 #include<string>
 6 #include<cstring>
 7 #include<algorithm>
 8 #include<functional>
 9 #include<iomanip>
10 #include<numeric>
11 #include<cmath>
12 #include<queue>
13 #include<vector>
14 #include<set>
15 #include<cctype>
16 const double PI = acos(-1.0);
17 const int INF = 0x3f3f3f3f;
18 const int NINF = -INF - 1;
19 const int maxn = 1e3 + 5;
20 typedef long long ll;
21 #define MOD 1000000007
22 using namespace std;
23 int dp[maxn][2];
24 struct node
25 
26     int x1, x2, h;
27     friend bool operator < (const node &m, const node &n)
28     
29         return m.h < n.h;
30     
31 a[maxn];
32 int main()
33 
34     int T;
35     scanf("%d", &T);
36     while (T--)
37     
38         int n, x, y, maxm;
39         scanf("%d %d %d %d", &n, &x, &y, &maxm);
40         for (int i = 0; i <= n + 1; ++i) dp[i][0] = dp[i][1] = INF;
41         for (int i = 1; i <= n; ++i)
42         
43             int p, q, c;
44             scanf("%d %d %d", &p, &q, &c);
45             a[i] = nodep, q, c;
46         
47         a[0] = node-20000, 20000, 0;
48         a[n + 1] = nodex, x, y;
49         sort(a, a + n + 1);
50 //        for (int i = 0; i <= n + 1; ++i) cout << a[i].x1 << ‘ ‘ << a[i].x2 << ‘ ‘ << a[i].h << endl;
51         for (int i = 1; i <= n + 1; ++i)
52         
53             int flag1 = 0, flag2 = 0;
54             for (int j = i - 1; j >= 0; --j)
55             
56                 if (a[i].h - a[j].h > maxm) break;
57                 if (flag1 && flag2) break;
58                 if (!flag1)
59                 
60                     if (j == 0)
61                     
62                         dp[i][0] = a[i].h;
63                         flag1++;
64                     
65                     else 
66                         if (a[j].x1 <= a[i].x1 && a[j].x2 >= a[i].x1)
67                         
68                             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);
69                             flag1++;
70                         
71                     
72                 
73                 if (!flag2)
74                 
75                     if (j == 0)
76                     
77                         dp[i][1] = a[i].h;
78                         flag2++;
79                     
80                     else 
81                         if (a[j].x1 <= a[i].x2 && a[j].x2 >= a[i].x2)
82                         
83                             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);
84                             flag2++;
85                         
86                     
87                 
88             
89 //            cout << i << ‘ ‘ << dp[i][0] << ‘ ‘ << dp[i][1] << endl;
90         
91 //        cout << dp[n + 1][0] << ‘ ‘ << dp[n + 1][1];
92         printf("%d\n", min(dp[n + 1][0], dp[n + 1][1]));
93     
94     return 0;
95