Finding Nemo
Time Limit:2000MS Memory Limit:30000K
Total Submit:1965 Accepted:370
Description
Nemo is a naughty boy. One day he went into the deep sea all by himself. Unfortunately, he became lost and couldn't find his way home. Therefore, he sent a signal to his father, Marlin, to ask for help.
After checking the map, Marlin found that the sea is like a labyrinth with walls and doors. All the walls are parallel to the X-axis or to the Y-axis. The thickness of the walls are assumed to be zero.
All the doors are opened on the walls and have a length of 1. Marlin cannot go through a wall unless there is a door on the wall. Because going through a door is dangerous (there may be some virulent medusas near the doors), Marlin wants to go through as few doors as he could to find Nemo.
Figure-1 shows an example of the labyrinth and the path Marlin went through to find Nemo.
We assume Marlin's initial position is at (0, 0). Given the position of Nemo and the configuration of walls and doors, please write a program to calculate the minimum number of doors Marlin has to go through in order to reach Nemo.
Input
The input consists of several test cases. Each test case is started by two non-negative integers M and N. M represents the number of walls in the labyrinth and N represents the number of doors.
Then follow M lines, each containing four integers that describe a wall in the following format:
x y d t
(x, y) indicates the lower-left point of the wall, d is the direction of the wall -- 0 means it's parallel to the X-axis and 1 means that it's parallel to the Y-axis, and t gives the length of the wall.
The coordinates of two ends of any wall will be in the range of [1,199].
Then there are N lines that give the description of the doors:
x y d
x, y, d have the same meaning as the walls. As the doors have fixed length of 1, t is omitted.
The last line of each case contains two positive float numbers:
f1 f2
(f1, f2) gives the position of Nemo. And it will not lie within any wall or door.
A test case of M = -1 and N = -1 indicates the end of input, and should not be processed.
Output
For each test case, in a separate line, please output the minimum number of doors Marlin has to go through in order to rescue his son. If he can't reach Nemo, output -1.
Sample Input
8 9
1 1 1 3
2 1 1 3
3 1 1 3
4 1 1 3
1 1 0 3
1 2 0 3
1 3 0 3
1 4 0 3
2 1 1
2 2 1
2 3 1
3 1 1
3 2 1
3 3 1
1 2 0
3 3 0
4 3 1
1.5 1.5
4 0
1 1 0 1
1 1 1 1
2 1 1 1
1 2 0 1
1.5 1.7
-1 -1
Sample Output
5
-1
第一次用priority_queue 发现效率还不错~
居然1y了 很是惊讶的说 rp突然好起来了？
我是从Nemo向Marlin搜索的 采用优先队列的方式就能找到最少的过门方式
细节部分还是要注意~~
Solution：
//by Optimistic
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
const int MAXINT = 2000000000;
const int N = 210;
struct Point{int x, y, p;}; //坐标，权值（经过门的个数）
bool wall[N][N][2], door[N][N][2], end[N][N], seted[N][N]; //墙壁，门，可直达点，标记
Point start; //起点（Nemo）
int nw, nd, xh, yh; //墙的个数，门的个数，x的最高值，y的最高值（限定搜索范围）
struct cmp
{
public: inline bool operator()(const Point& a, const Point& b) {return a.p > b.p;}
};
priority_queue <Point, vector<Point>, cmp> pq; //优先队列
void set_end(int x, int y) //设置可直达点(不通过任何门)
{
end[x][y] = 1;
if(x-1 <= xh && x-1 >= 0 && y <= yh && y >= 0 && wall[x][y][1] == 0
&& door[x][y][1] == 0 && !end[x-1][y]) set_end(x-1, y);
if(x <= xh && x >= 0 && y-1 <= yh && y-1 >= 0 && wall[x][y][0] == 0
&& door[x][y][0] == 0 && !end[x][y-1]) set_end(x, y-1);
if(x+1 <= xh && x+1 >= 0 && y <= yh && y >= 0 && wall[x+1][y][1] == 0
&& door[x+1][y][1] == 0 && !end[x+1][y]) set_end(x+1, y);
if(x <= xh && x >= 0 && y+1 <= yh && y+1 >= 0 && wall[x][y+1][0] == 0
&& door[x][y+1][0] == 0 && !end[x][y+1]) set_end(x, y+1);
}
void init()
{
int i, x, y, d, t, j;
double dx, dy;
memset(wall, 0, sizeof(wall));
memset(door, 0, sizeof(door));
memset(end, 0, sizeof(end));
yh = -MAXINT; xh = -MAXINT;
for(i = 0; i<nw; i++) {
scanf("%d%d%d%d", &x, &y, &d, &t);
if(x > xh) xh = x;
if(y > yh) yh = y;
if(d == 0) {
for(j = 0; j<t; j++) {
wall[x+j][y][0] = 1;
}
}
else {
for(j = 0; j<t; j++) {
wall[x][y+j][1] = 1;
}
}
}
for(i = 0; i<nd; i++) {
scanf("%d%d%d", &x, &y, &d);
if(x > xh) xh = x;
if(y > yh) yh = y;
door[x][y][d] = 1;
wall[x][y][d] = 0; //门要把墙覆盖 sample中可以看出来
}
cin >> dx >> dy;
start.x = (int)dx, start.y = (int)dy, start.p = 0; //设置起点 初始化权值为0
memset(seted, 0, sizeof(seted));
set_end(0, 0); //设置可直达点(不通过任何门)
yh++; xh++;
}
void work()
{
if(start.x < 0 || start.x >= 200 || start.y < 0 || start.y >= 200 || xh <0 || yh <0) //特殊情况
{ printf("0\n"); return ; }
memset(seted, 0, sizeof(seted));
while(!pq.empty()) pq.pop();
pq.push(start); //起点入列
seted[start.x][start.y] = 1;
while(!pq.empty()) {
Point cur = pq.top(); //取权值最大的点
pq.pop();
if(end[cur.x][cur.y] == 1) { printf("%d\n", cur.p); return;} //找到解
int x= cur.x, y = cur.y;
Point now;
if(x-1 <= xh && x-1 >= 0 && y <= yh && y >= 0 && wall[x][y][1] == 0 && !seted[x-1][y]) { //左搜索
if(door[x][y][1] == 1) now.p = cur.p + 1;
else now.p = cur.p;
now.x = x -1; now.y = y;
pq.push(now);
}
if(x <= xh && x >= 0 && y - 1 <= yh && y-1 >= 0 && wall[x][y][0] == 0 && !seted[x][y-1]) { //下搜索
if(door[x][y][0] == 1) now.p = cur.p + 1;
else now.p = cur.p;
now.x = x; now.y = y-1;
pq.push(now);
seted[now.x][now.y] = 1;
}
if(x+1 <= xh && x+1 >= 0 && y <= yh && y >= 0 && wall[x+1][y][1] == 0 && !seted[x+1][y]) { //右搜索
if(door[x+1][y][1] == 1) now.p = cur.p + 1;
else now.p = cur.p;
now.x = x + 1; now.y = y;
pq.push(now);
seted[now.x][now.y] = 1;
}
if(x <= xh && x >= 0 && y+1 <= yh && y+1 >= 0 && wall[x][y+1][0] == 0 && !seted[x][y+1]) { //上搜索
if(door[x][y+1][0] == 1) now.p = cur.p + 1;
else now.p = cur.p;
now.x = x; now.y = y+1;
pq.push(now);
seted[now.x][now.y] = 1;
}
}
printf("-1\n"); //无解
}
int main()
{
while(scanf("%d %d", &nw, &nd), !(nw == -1 && nd == -1)) {
init();
work();
}
return 0;
}