Finding Nemo
| Time Limit: 2000MS | Memory Limit: 30000K | |
| Total Submissions: 7902 | Accepted: 1827 |
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
题意：有m个墙，n个门。接下来m行 每行输入x，y，d，t分别代表左下角的(X,Y)坐标，d=1代表代表墙平行于y轴，d=0代表墙平行于x轴，t代表该行或该列上有多少墙。
思路：每个网格的坐标用网格左下角坐标来代替，用第三维来代表网格的上边和右边。然后BFS搜索，要全部搜索完取最小值。
三维map的结果0代表空地，1代表墙，2代表门，map[X][Y][0]代表此表格的上方，map[X][Y][1]代表此表格的右方。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int inf=0x3f3f3f3f;
struct node {
int x, y;
int ans;
} f1,f2;
int map[210][210][3];
int vis[210][210];
int jx[]= {0,0,1,-1};
int jy[]= {1,-1,0,0};
int min1;
void bfs(int s, int e)
{
int i, j;
queue<node>q;
f1.x=s;
f1.y=e;
f1.ans=0;
vis[s][e]=1;
q.push(f1);
min1=inf;
while(!q.empty()) {
f1=q.front();
q.pop();
if(f1.x<=0||f1.x>=199||f1.y<=0||f1.y>=199) {//边界处理，但是不能跳出，只能跳过，因为存在多个门的情况
min1=min(min1,f1.ans); //走出去之后，保存最小通过门数
continue ;
}
for(i=0; i<4; i++) {
f2.x=f1.x+jx[i];
f2.y=f1.y+jy[i];
if(i==0) {//向上走
if(!vis[f2.x][f2.y]&&map[f1.x][f1.y][0]!=2) {
if(map[f1.x][f1.y][0]==1)
f2.ans=f1.ans+1;
else
f2.ans=f1.ans;
vis[f2.x][f2.y]=1;
q.push(f2);
}
} else if(i==1) {//向下走
if(!vis[f2.x][f2.y]&&map[f2.x][f2.y][0]!=2) {
if(map[f2.x][f2.y][0]==1)
f2.ans=f1.ans+1;
else
f2.ans=f1.ans;
vis[f2.x][f2.y]=1;
q.push(f2);
}
} else if(i==2) {//向右走
if(!vis[f2.x][f2.y]&&map[f1.x][f1.y][1]!=2) {
if(map[f1.x][f1.y][1]==1)
f2.ans=f1.ans+1;
else
f2.ans=f1.ans;
vis[f2.x][f2.y]=1;
q.push(f2);
}
} else if(i==3) {//向左走
if(!vis[f2.x][f2.y]&&map[f2.x][f2.y][1]!=2) {
if(map[f2.x][f2.y][1]==1)
f2.ans=f1.ans+1;
else
f2.ans=f1.ans;
vis[f2.x][f2.y]=1;
q.push(f2);
}
}
}
}
}
int main()
{
int n, m, i, j;
int x,y,d,t;
double a1,a2;
int int_x,int_y;
while(scanf("%d %d",&m,&n)!=EOF) {
if(n==-1&&m==-1) break;
memset(map,0,sizeof(map));
memset(vis,0,sizeof(vis));
for(i=0; i<m; i++) {
scanf("%d %d %d %d",&x, &y, &d, &t);
if(d) {
for(j=0; j<t; j++) {
map[x-1][y+j][1]=2;
}
} else {
for(j=0; j<t; j++) {
map[x+j][y-1][0]=2;
}
}
}
for(i=0; i<n; i++) {
scanf("%d %d %d",&x,&y,&d);
if(d) {
map[x-1][y][1]=1;
} else {
map[x][y-1][0]=1;
}
}
scanf("%lf %lf",&a1,&a2);
int_x=a1;
int_y=a2;
if(int_x<=0||int_x>=199||int_y<=0||int_y>=199) {//后台可能出现 0,0 200+,200+，虽然不符合题目描述
printf("0\n");
continue ;
}
bfs(int_x,int_y);
if(min1==inf)
printf("-1\n");
else
printf("%d\n",min1);
}
return 0;
}