HDU 5336 XYZ and Drops

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 
    
    
    
    
    
    
    
     T  seconds, 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
    
    
    
    
 
    
    
    
    
来个优先级队列记录一下时间，暴力的搞吧
    
    
    
    

    
    
    
    
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<algorithm>
#include<string>
#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
using namespace std;
const ll maxn=105;
int T,n,m,t,f[maxn][maxn],x,y,a[maxn],b[maxn],c[maxn][maxn];

struct point
{
    int x,y,z,t,now;
    point(){}
    point(int x,int y,int z,int t,int now):
    x(x),y(y),z(z),t(t),now(now) {};
    bool operator <(const point &a) const
    {
        return now+t>a.now+a.t;
    }
};

int get(int x,int y,int d)
{
    if (d&1)
    {
        if (d==1)
        {
            for (int i=y+1;i<=m;i++)
                if (f[x][i]) return i-y;
            return 0;
        }
        else 
        {
            for (int i=y-1;i>=1;i--)
                if (f[x][i]) return y-i;
            return 0;
        }
    }
    else 
    {
        if (d==0) 
        {
            for (int i=x-1;i>=1;i--)
                if (f[i][y]) return x-i;
            return 0;
        }
        else 
        {
            for (int i=x+1;i<=n;i++)
                if (f[i][y]) return i-x;
            return 0;
        }
    }
}

int main()
{
    //scanf("%d",&T);
    while (~scanf("%d%d%d%d",&n,&m,&t,&T))
    {
        memset(f,0,sizeof(f));
        memset(c,0,sizeof(c));
        for (int i=1;i<=t;i++)
        {
            scanf("%d%d",&x,&y);
            a[i]=x;    b[i]=y;
            scanf("%d",&f[x][y]);
        }
        scanf("%d%d",&x,&y);
        priority_queue<point> p;
        for (int i=0;i<4;i++) 
        {
            int k=get(x,y,i);
            if (k>0) p.push(point(x,y,i,k,0));
        }
        while (!p.empty())
        {
            point tp,q=p.top();    p.pop();
            
            for (;;p.pop())
            {
                if (p.empty()) break;
                tp=p.top();
                if(tp.now+tp.t!=q.now+q.t) break;
                int k=get(tp.x,tp.y,tp.z);
                if (!k) continue;
                if (k!=tp.t) p.push(point(tp.x,tp.y,tp.z,k,tp.now));
                else 
               {
                if (k+tp.now>T) break;
                if (tp.z==0) x=tp.x-k,y=tp.y;
                if (tp.z==1) x=tp.x,y=tp.y+k;
                if (tp.z==2) x=tp.x+k,y=tp.y;
                if (tp.z==3) x=tp.x,y=tp.y-k;
                f[x][y]++;
                }
            }
            int k=get(q.x,q.y,q.z);
            if (k) 
            if (k!=q.t) p.push(point(q.x,q.y,q.z,k,q.now));
            else 
            {
                if (k+q.now>T) break;
                if (q.z==0) x=q.x-k,y=q.y;
                if (q.z==1) x=q.x,y=q.y+k;
                if (q.z==2) x=q.x+k,y=q.y;
                if (q.z==3) x=q.x,y=q.y-k;
                f[x][y]++;
            }
            for(int i=1;i<=t;i++)
            {
                x=a[i];    y=b[i];
                if (f[x][y]>4)
                {
                    f[x][y]=0;
                    c[x][y]=q.t+q.now;
                    for (int j=0;j<4;j++) 
                    {
                        int u=get(x,y,j);
                        if (u>0) p.push(point(x,y,j,u,q.t+q.now));
                    }
                }
            }
        }
        for (int i=1;i<=t;i++)
        {
            if (f[a[i]][b[i]]) printf("1 %d\n",f[a[i]][b[i]]);
            else printf("0 %d\n",c[a[i]][b[i]]);
        }
    }
    return 0;
}