XYZ and Drops
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1181 Accepted Submission(s): 372
Problem Description
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
T seconds, can you help him?
1≤r≤100,
1≤c≤100,
1≤n≤100,
1≤T≤10000
Input
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
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
Source
2015 Multi-University Training Contest 4
点击打开链接
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<queue>
using namespace std;
int r,c,n,pt;
int mp[101][101];
struct node
{
int xx;
int yy;
int tt;
int ll;
int dd;
bool operator < (const node &px) const {
return tt>px.tt;//最小值优先
}
};
void BFS(int x,int y)
{
priority_queue<node>q;
while(!q.empty())
{
q.pop();
}
struct node t,f;
for(int i=-1; i<=1; i++)
{
for(int j=-1; j<=1; j++)
{
if(i!=j && (abs(i-j)<=1))
{
t.xx = x;
t.yy = y;
t.tt = 0;
t.dd = i;
t.ll = j;
if(t.xx<=r && t.yy<=c && t.xx>0 && t.yy>0)
{
q.push(t);
}
}
}
}
while(!q.empty())
{
t = q.top();
q.pop();
f.xx = t.xx + t.dd;
f.yy = t.yy + t.ll;
f.tt = t.tt + 1;
if(f.xx<=r && f.yy<=c && f.xx>0 && f.yy>0)
{
if(mp[f.xx][f.yy]>0 && f.tt<=pt)
{
mp[f.xx][f.yy] += 1;
}
else
{
if(mp[f.xx][f.yy]<0 && f.tt == -mp[f.xx][f.yy])
{
}
else if(f.tt<=pt)
{
f.ll = t.ll;
f.dd = t.dd;
q.push(f);
}
}
if(mp[f.xx][f.yy]>4)
{
mp[f.xx][f.yy] = -1 * f.tt;
for(int i=-1; i<=1; i++)
{
for(int j=-1; j<=1; j++)
{
if(i!=j && (abs(i-j)<=1))
{
struct node aa;
aa.xx = f.xx;
aa.yy = f.yy;
aa.tt = f.tt;
aa.ll = j;
aa.dd = i;
if(aa.xx<=r && aa.yy<=c && aa.xx>0 && aa.yy>0 && aa.tt<=pt)
{
q.push(aa);
}
}
}
}
}
}
}
}
int main()
{
while(scanf("%d%d%d%d",&r,&c,&n,&pt)!=EOF)
{
int pl[110],pr[110];
int x,y,z;
memset(mp,0,sizeof(mp));
for(int i=0; i<n; i++)
{
scanf("%d%d%d",&x,&y,&z);
pl[i] = x;
pr[i] = y;
mp[x][y] = z;
}
scanf("%d%d",&x,&y);
BFS(x,y);
for(int i=0; i<n; i++)
{
if(mp[pl[i]][pr[i]]>0)
{
printf("1 %d\n",mp[pl[i]][pr[i]]);
}
else if(mp[pl[i]][pr[i]]<0)
{
printf("0 %d\n",-mp[pl[i]][pr[i]]);
}
}
}
return 0;
}
