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
第一种方法是枚举的时间，效率较低，比赛时跑了265ms。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#define ll long long
using namespace std;
int jl[1110];
int r,c,n,t;
struct node
{
int x,y,d;
};
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int Map[110][110];
struct node2
{
int water,time;
};
node2 ans [110][110];
int in(int x,int y)
{
if(x>=1&&x<=r&&y>=1&&y<=c)
return 1;
return 0;
}
queue<node> q;
queue<node> q2;
int bfs( )
{
node next,tmp,fen;
int xx,yy;
for(int ttt = 0; ttt<=t; ttt++)
{
// for(int i=1; i<=r; i++)
// {
// for(int j=1; j<=c; j++)
// {
// cout<<Map[i][j]<<" ";
// }
// cout<<endl;
// }
// cout<<endl;
for(int i=0; i<n; i++)
{
xx = jl[i]/1000, yy = jl[i]%1000;
if(Map[xx][yy]>4)
{
Map[xx][yy] = 0;
ans[xx][yy].water = 0;
ans[xx][yy].time = ttt;
for(int j=0; j<4; j++)
{
fen.x = xx,fen.y = yy,fen.d = j;
q.push(fen);
}
}
else
{
ans[xx][yy].water = Map[xx][yy];
}
}
while(!q.empty())
{
node tmp = q.front();
q.pop();
int xx = tmp.x + dir[tmp.d][0];
int yy = tmp.y + dir[tmp.d][1];
if(!in(xx,yy))
continue;
if(Map[xx][yy]==0)
{
next.x = xx;
next.y = yy;
next.d = tmp.d;
q2.push(next);
}
else
Map[xx][yy] ++;
}
while(!q2.empty())
{
q.push(q2.front());
q2.pop();
}
}
return 0;
}
int main()
{
while(cin>>r>>c>>n>>t)
{
memset(Map,0,sizeof(Map));
for(int i=0; i<n; i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
jl[i] = x*1000+y;
ans[x][y].water = z;
ans[x][y].time = 0;
Map[x][y] += z;
}
int sx,sy;
cin>>sx>>sy;
while(!q.empty())
q.pop();
while(!q2.empty())
q.pop();
node tmp;
for(int i=0; i<4; i++)
{
tmp.x = sx,tmp.y = sy;
tmp.d = i;
q.push(tmp);
}
bfs();
int ex,ey;
node2 flag ;
for(int i=0; i<n; i++)
{
ex = jl[i]/1000;
ey = jl[i]%1000;
flag = ans[ex][ey];
if(flag.water == 0)
cout<<0<<" "<<flag.time<<endl;
else
cout<<1<<" "<<flag.water<<endl;
}
}
return 0;
}
第二种方法只需把当前这一秒所有的小水珠都处理完再判断是否爆裂。赛后跑了46ms。
据说是赛前该小了数据，要不然第一种方法就会T掉。
</pre><pre name="code" class="cpp">#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#define ll long long
using namespace std;
int jl[1110];
int r,c,n,t;
struct node
{
int x,y,d;
int mov;
};
int dir[4][2] = {{0,1},{0,-1},{1,0},{-1,0}};
int Map[110][110];
struct node2
{
int water,time;
};
node2 ans [110][110];
int in(int x,int y)
{
if(x>=1&&x<=r&&y>=1&&y<=c)
return 1;
return 0;
}
queue<node> q;
int bfs( )
{
node next,tmp,fen;
int xx,yy;
while(!q.empty()&&q.front().mov<t)
{
node tmp = q.front();
q.pop();
int xx = tmp.x + dir[tmp.d][0];
int yy = tmp.y + dir[tmp.d][1];
// if(in(xx,yy)) continue;
//如果后面还有判断条件 ，不要这样写
if(in(xx,yy))
{
if(Map[xx][yy]==0)
{
next.x = xx;
next.y = yy;
next.d = tmp.d;
next.mov = tmp.mov + 1;
q.push(next);
}
else
Map[xx][yy] ++;
}
if(q.empty()||q.front().mov != tmp.mov)
{
for(int i=0; i<n; ++i)
{
xx = jl[i]/1000, yy = jl[i]%1000;
if(Map[xx][yy]>4)
{
Map[xx][yy] = 0;
ans[xx][yy].water = 0;
ans[xx][yy].time = tmp.mov + 1;
for(int j=0; j<4; ++j)
{
fen.x = xx,fen.y = yy,fen.d = j;
fen.mov = tmp.mov + 1;
q.push(fen);
}
}
else
ans[xx][yy].water = Map[xx][yy];
}
}
}
return 0;
}
int main()
{
while(scanf("%d%d%d%d",&r,&c,&n,&t)!=EOF)
{
memset(Map,0,sizeof(Map));
for(int i=0; i<n; ++i)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
jl[i] = x*1000+y;
ans[x][y].water = z;
ans[x][y].time = 0;
Map[x][y] += z;
}
int sx,sy;
scanf("%d%d",&sx,&sy);
while(!q.empty())
q.pop();
node tmp;
for(int i=0; i<4; ++i)
{
tmp.x = sx,tmp.y = sy;
tmp.d = i;
tmp.mov = 0;
q.push(tmp);
}
bfs();
int ex,ey;
node2 flag ;
for(int i=0; i<n; ++i)
{
ex = jl[i]/1000;
ey = jl[i]%1000;
flag = ans[ex][ey];
if(flag.water == 0)
printf("0 %d\n",flag.time);
else
printf("1 %d\n",flag.water);
}
}
return 0;
}
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