标签：
Time Limit: 1000MS Memory Limit: 65536K
|
Description |
题目描述 |
|
Bessie hears that an extraordinary meteor shower is coming; reports say that these meteors will crash into earth and destroy anything they hit. Anxious for her safety, she vows to find her way to a safe location (one that is never destroyed by a meteor) . She is currently grazing at the origin in the coordinate plane and wants to move to a new, safer location while avoiding being destroyed by meteors along her way.
The reports say that M meteors (1 ≤ M ≤ 50,000) will strike, with meteor i will striking point (Xi, Yi) (0 ≤ Xi ≤ 300; 0 ≤ Yi ≤ 300) at time Ti (0 ≤ Ti ≤ 1,000). Each meteor destroys the point that it strikes and also the four rectilinearly adjacent lattice points.
Bessie leaves the origin at time 0 and can travel in the first quadrant and parallel to the axes at the rate of one distance unit per second to any of the (often 4) adjacent rectilinear points that are not yet destroyed by a meteor. She cannot be located on a point at any time greater than or equal to the time it is destroyed).
Determine the minimum time it takes Bessie to get to a safe place. |
Bessie听说有场史无前例的流星雨即将来临；有谶言：陨星将落，徒留灰烬。为保生机，她誓将找寻安全之所（永避星坠之地）。目前她正在平面坐标系的原点放牧，打算在群星断其生路前转移至安全地点。
此次共有M (1 ≤ M ≤ 50,000)颗流星来袭，流星i将在时间点Ti (0 ≤ Ti ≤ 1,000) 袭击点 (Xi, Yi) (0 ≤ Xi ≤ 300; 0 ≤ Yi ≤ 300)。每颗流星都将摧毁落点及其相邻四点的区域。
Bessie在0时刻时处于原点，且只能行于第一象限，以平行与坐标轴每秒一个单位长度的速度奔走于未被毁坏的相邻（通常为4）点上。在某点被摧毁的刹那及其往后的时刻，她都无法进入该点。 |
|
Input |
输入 |
|
* Line 1: A single integer: M * Lines 2..M+1: Line i+1 contains three space-separated integers: Xi, Yi, and Ti |
* 第1行: 一个整数: M * 第2..M+1行: 第i+1行包含由空格分隔的三个整数: Xi, Yi, and Ti |
|
Output |
输出 |
|
* Line 1: The minimum time it takes Bessie to get to a safe place or -1 if it is impossible. |
* 仅一行: Bessie寻得安全点所花费的最短时间，无解则为-1。 |
|
Sample Input - 输入样例 |
Sample Output - 输出样例 |
4
0 0 2
2 1 2
1 1 2
0 3 5
|
5
|
【题解】
从起点开始进行SPFA（DFS）即可。
需要注意的是在读取各个落点的数据时，只保留最早被毁灭的时间点。
【代码 C++】
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <queue> 5 #define mx 305 6 int dg[mx][mx], tim[mx][mx], drc[5][2] = { 1, 0, -1, 0, 0, 1, 0, -1, 0, 0 }; 7 struct Point{ 8 int y, x; 9 }temp, nxt; 10 int main(){ 11 int m, x, y, t, i, nowT, opt; 12 scanf("%d", &m); 13 memset(dg, -1, sizeof(dg)); 14 while (m--){ 15 scanf("%d%d%d", &x, &y, &t); 16 ++y; ++x; 17 for (i = 0; i < 5; ++i){ 18 temp.y = y + drc[i][1]; temp.x = x + drc[i][0]; 19 if (~dg[temp.y][temp.x]) dg[temp.y][temp.x] = std::min(dg[temp.y][temp.x], t); 20 else dg[temp.y][temp.x] = t; 21 } 22 } 23 24 memset(dg, 0, sizeof(dg[0])); memset(dg[mx - 1], 0, sizeof(dg[0])); 25 for (i = 0; i < mx; ++i) dg[i][0] = dg[i][mx - 1] = 0; 26 memset(tim, 127, sizeof(tim)); 27 nowT = 0; opt = 2100000000; 28 std::queue<Point> q; 29 tim[1][1] = 0; temp.y = temp.x = 1; q.push(temp); 30 while (!q.empty()){ 31 temp = q.front(); q.pop(); 32 if (dg[temp.y][temp.x] == -1){ 33 opt = std::min(opt, tim[temp.y][temp.x]); 34 continue; 35 } 36 for (i = 0; i < 4; ++i){ 37 nxt.y = temp.y + drc[i][0]; nxt.x = temp.x + drc[i][1]; 38 if (tim[temp.y][temp.x] + 1 < dg[nxt.y][nxt.x] || dg[nxt.y][nxt.x] == -1){ 39 if (tim[temp.y][temp.x] + 1 < tim[nxt.y][nxt.x]){ 40 tim[nxt.y][nxt.x] = tim[temp.y][temp.x] + 1; 41 q.push(nxt); 42 } 43 } 44 } 45 } 46 if (opt == 2100000000) puts("-1"); 47 else printf("%d", opt); 48 return 0; 49 }
标签：
原文地址：http://www.cnblogs.com/Simon-X/p/5891221.html
