pku 1661 Help Jimmy

#include <iostream> #include <climits> #include <algorithm> using namespace std; struct Board { int left, right; int height; }; Board board[1005]; int Left[1005], Right[1005]; int n; int N, X, Y, MAX; #define INTMAX INT_MAX/10 bool cmp(const Board& lhs, const Board& rhs) { if(lhs.height > rhs.height) return true; else if(lhs.height == rhs.height && lhs.left < rhs.left) return true; else return false; } inline int min(const int& lhs, const int& rhs) { return lhs < rhs ? lhs:rhs; } inline bool notBeCovered(int p, int i, int j) { for(int k = i-1; k >= j+1; --k) if(board[k].left <= p && board[k].right >= p) return false; return true; } inline bool canGoToGroud(int p, int i) { for(int k = i+1; k <= N; k++) if(board[k].left <= p && board[k].right >= p) return false; return true; } int main() { scanf("%d", &n); while(n--) { scanf("%d%d%d%d", &N, &X, &Y, &MAX); for(int i = 1; i <= N; ++i) scanf("%d%d%d", &board[i].left, &board[i].right, &board[i].height); sort(board+1, board+N+1, cmp); memset(Left, 0x7777, sizeof(Left)); memset(Right, 0x7777, sizeof(Right)); board[0].left = board[0].right = X; board[0].height = Y; Left[0] = Right[0] = 0; for(int i = 1; i <= N; ++i) { int min1 = INTMAX, min2 = INTMAX; for(int j = i-1; j >= 0; j--) if(board[j].height - board[i].height > 0 && board[j].height - board[i].height <= MAX) { int temp1 = INTMAX; if(board[j].left <= board[i].right&&board[j].left >= board[i].left && notBeCovered(board[j].left, i, j)) temp1 = min(temp1, board[j].left-board[i].left+Left[j]); if(board[j].right <= board[i].right&&board[j].right >= board[i].left && notBeCovered(board[j].right, i, j)) temp1 = min(temp1, board[j].right-board[i].left+Right[j]); if(temp1 < min1) min1 = temp1; int temp2 = INTMAX; if(board[j].left <= board[i].right&&board[j].left >= board[i].left && notBeCovered(board[j].left, i, j)) temp2 = min(temp2, board[i].right-board[j].left+Left[j]); if(board[j].right <= board[i].right&&board[j].right >= board[i].left && notBeCovered(board[j].right, i, j)) temp2 = min(temp2, board[i].right-board[j].right+Right[j]); if(temp2 < min2) min2 = temp2; } Left[i] = min1, Right[i] = min2; } int Min = INTMAX; for (int i = 0; i <= N; ++i) { if(board[i].height <= MAX) { if(canGoToGroud(board[i].left, i) && Left[i] < Min) Min = Left[i]; if(canGoToGroud(board[i].right, i) && Right[i] < Min) Min = Right[i]; } } printf("%d/n", Min+Y); } return 0; } /* 思路相对比较简单: 设到达第i个板左端的最短时间Left[i], 到达第i个板右端的最短时间Right[i]; 所求即为每条可以到达地面的板(包括初始位置)的最小值 (其中要刨去超过MAX的). //此题需要注意的地方颇多: 1. INTMAX 不可取INT_MAX, 否则INT_MAX再加上一个数可能溢出变成负数, 以后做题时也一定要注意!!! 2. 注意Jimmy可以直接到地面的情形, 即: for (int i = 1; i <= N; ++i) => for (int i = 0; i <= N; ++i) 3.要判断Jimmy初始位置能否到达该板: for(int i = 1; i <= N; ++i) { int min1 = INTMAX, min2 = INTMAX; for(int j = i-1; j >= 0; j--) ... } 4. 判断每个板是否能否直接到地面时, 不要忘记高度限制, 我就在这WA了2次: for (int i = 0; i <= N; ++i) { if(board[i].height <= MAX) .... } */