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] = node{p, q, c}; 46 } 47 a[0] = node{-20000, 20000, 0}; 48 a[n + 1] = node{x, 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 }