HDOJ 5336 XYZ and Drops 模拟

暴力模拟

XYZ and Drops

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

Total Submission(s): 230    Accepted Submission(s): 35


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

 

Source

2015 Multi-University Training Contest 4

 

/* ***********************************************
Author :CKboss
Created Time :2015年07月30日 星期四 14时52分47秒
File Name :1010.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int dir_x[4]={0,0,-1,1};
const int dir_y[4]={1,-1,0,0};

int r,c,n,T;

int mid[111][111];
int SZ[111],Time[111];

int X[111],Y[111];

void init()
{
memset(SZ,0,sizeof(SZ));
memset(mid,0,sizeof(mid));
memset(Time,0,sizeof(Time));
}

struct Node
{
int x,y,d,t;

void toString()
{
printf("(%d,%d) d: %d t: %d\n",x,y,d,t);
}
};

bool isInside(int x,int y)
{
if((x>=1&&x<=r)&&(y>=1&&y<=c)) return true;
return false;
}

void solve(int _x,int _y)
{
queue<Node> q;

for(int i=0;i<4;i++)
{
Node nd;
nd.x=_x; nd.y=_y; nd.d=i; nd.t=0;
q.push(nd);
}

while(!q.empty())
{
Node u=q.front(); q.pop();

//u.toString();

/// boom waterdrop number
int bbb=-1;

int d=u.d;
Node v;

v.x=u.x+dir_x[d];
v.y=u.y+dir_y[d];
v.d=u.d;
v.t=u.t+1;

/// is in side
if(isInside(v.x,v.y)==false) continue;

int id=mid[v.x][v.y];

/// zhe waterdrop is boomming
if(id==0)
{
if(v.t<T)
{
q.push(v);
}
}
else
{
if(Time[id]!=0)
{
if(Time[id]>=v.t) continue;
else
{
if(v.t<T)
{
q.push(v);
}
}
}
/// meet with waterdrop
else if(Time[id]==0)
{
SZ[id]++;
if(SZ[id]>4)
{
Time[id]=v.t;
//mid[v.x][v.y]=0;
bbb=id;
}
}
}

if(bbb!=-1)
{
/// add 4 new node
for(int j=0;j<4;j++)
{
Node n1;
n1.x=X[bbb]; n1.y=Y[bbb];
n1.d=j; n1.t=v.t;

if(n1.t<T)
{
q.push(n1);
}
}
}
}

//cout<<".......end............\n";
}

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);

while(scanf("%d%d%d%d",&r,&c,&n,&T)!=EOF)
{
init();
for(int i=1,s;i<=n;i++)
{
scanf("%d%d%d",X+i,Y+i,&s);
int x=X[i],y=Y[i];
mid[x][y]=i;
SZ[i]=s;
}

int x,y;
scanf("%d%d",&x,&y);
solve(x,y);

for(int i=1;i<=n;i++)
{
if(Time[i]!=0) printf("%d %d\n",0,Time[i]);
else printf("%d %d\n",1,SZ[i]);
}
}

return 0;
}