HDU 5336 XYZ and Drops(bfs)

XYZ and Drops

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 736    Accepted Submission(s): 203


Problem Description

XYZ is playing an interesting game called "drops". It is played on a r∗c grid.
Each grid cell is either empty, or occupied by a waterdrop. Each waterdrop has a property "size". The waterdrop cracks when its size is larger than 4, and produces 4 small drops moving towards 4 different directions (up, down, left and right). 

In every second, every small drop moves to the next cell of its direction. It is possible that multiple small drops can be at same cell, and they won't collide. Then for each cell occupied by a waterdrop, the waterdrop's size increases by the number of the
small drops in this cell, and these small drops disappears. 

You are given a game and a position (x, y),
before the first second there is a waterdrop cracking at position (x, y).
XYZ wants to know each waterdrop's status after Tseconds,
can you help him?

1≤r≤100, 1≤c≤100, 1≤n≤100, 1≤T≤10000

 

Input

The first line contains four integers r, c, n and T. n stands
for the numbers of waterdrops at the beginning. 

Each line of the following n lines
contains three integers xi, yi, sizei,
meaning that the i-th
waterdrop is at position (xi, yi)
and its size is sizei.
(1≤sizei≤4)

The next line contains two integers x, y. 

It is guaranteed that all the positions in the input are distinct. 

Multiple test cases (about 100 cases), please read until EOF (End Of File).

 

Output

n lines.
Each line contains two integers Ai, Bi: 

If the i-th
waterdrop cracks in T seconds, Ai=0, Bi= the
time when it cracked. 

If the i-th
waterdrop doesn't crack in T seconds, Ai=1, Bi= its
size after T seconds.

 

Sample Input

4 4 5 10
2 1 4
2 3 3
2 4 4
3 1 2
4 3 4
4 4

 

Sample Output

0 5
0 3
0 2
1 3
0 1

 

Author

XJZX

 
直接代码：

#include<cstdio>
#include<cstring>
#include<vector>
#include<iostream>
#include <algorithm>
#include <queue>
using namespace std;

int Size[110][110];
bool vis[110][110];
int r, c, n ,T;
int dir[4][2] = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
int dx[10010], dy[10010];
int time[110][110];
int crash[110][110];
struct P
{
int x, y, dis, t;
}now;
queue <P> ac;
bool ok(int x, int y)
{
if(x >= 1 && x <= r && y >= 1 && y <= c)
return true;
else
return false;
}
int main()
{
while(~scanf("%d%d%d%d", &r, &c, &n, &T))
{
memset(Size, 0, sizeof(Size));
memset(vis, 0, sizeof(vis));
memset(crash, -1, sizeof(crash));
for(int i=0; i<n; i++)
{
scanf("%d%d", &dx[i], &dy[i]);
scanf("%d", &Size[dx[i]][dy[i]]);
vis[dx[i]][dy[i]] = true;
}
int sx, sy;
scanf("%d%d", &sx, &sy);
Size[sx][sy] = 5;
vis[sx][sy] = true;
ac.push((P){sx, sy, 0, 0});
while(!ac.empty())
{
now = ac.front();
ac.pop();
int x = now.x, y = now.y, dis = now.dis, t = now.t;
if(crash[x][y] == t) continue;
if(vis[x][y])
{
Size[x][y]++;
if(Size[x][y] > 4)
{
crash[x][y] = t;
vis[x][y] = false;
time[x][y] = t;
t = t + 1;
for(int i = 0; i < 4; i++)
{
int xx = x + dir[i][0], yy = y + dir[i][1];
if(ok(xx, yy) && t <= T)
{
ac.push((P){xx, yy, i, t});
}
}
}
}
else
{
x = x + dir[dis][0], y = y + dir[dis][1], t = t + 1;
if(ok(x, y) && t <= T)
ac.push((P){x, y, dis, t});
}
}
for(int i=0; i<n; i++)
{
if(vis[dx[i]][dy[i]])
{
printf("1 %d\n", Size[dx[i]][dy[i]]);
}
else
{
printf("0 %d\n", time[dx[i]][dy[i]]);
}
}
}
return 0;
}