XYZ and Drops (hdu 5336 bfs)

XYZ and Drops

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

Total Submission(s): 588 Accepted Submission(s): 157



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

Author

XJZX



Source

2015 Multi-University Training Contest 4



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wange2014 | We have carefully selected several similar problems for you: 5338 5337 5335 5334 5333



题意：n个大水滴在r*c的平面上，（x，y）处有一水滴分裂，分成四个小水滴向四个方向前进，其他n个大水滴的初始大小为1~4，若大水滴被小水滴撞到大水滴大小增加一，当大水滴大小超过四时会分裂，同样向四个方向，这样连锁反应，问最后T时刻n个水滴的状态。

思路：bfs，比赛写的时候一个小bug没看出来，思路上的一点漏洞，遗憾。

代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#define DBG printf("Hi\n")
using namespace std;

#define mod 10000007

#define N 100005

typedef __int64 ll;

struct Node
{
    int x,y,t,d;
    bool operator<(const Node &a)const
    {
        return t>a.t;
    }
};

int r,c,n,T;
int num[111][111];//该点大水滴的size
int id[111][111];
int q[111][2];
bool have[111][111];
int out[111][111];
int dir[4][2]={1,0,0,1,-1,0,0,-1};

bool isok(int x,int y)
{
    if (x>=0&&x<r&&y>=0&&y<c) return true;
    return false;
}

int Search(int d,int x,int y,int &X,int &Y)
{
    int ans=0;
    while (isok(x,y)&&!have[x][y])
    {
        x=x+dir[d][0];
        y=y+dir[d][1];
        ans++;
    }
    if (!isok(x,y)) return -1;
    X=x;Y=y;
    return ans+1;
}

void bfs(int x,int y)
{
    Node st,now;
    priority_queue<Node>Q;
    while (!Q.empty()) Q.pop();
    st.x=x;st.y=y;
    st.t=0;
    Q.push(st);
    int cnt=0;
    while (!Q.empty())
    {
        if (cnt>=n+1) break;
        st=Q.top();
        Q.pop();
        if (!have[st.x][st.y]) 
        {   //比赛时这个大括号内的没写，WA
            if (out[st.x][st.y]==st.t) continue;
            int ss=Search(st.d,st.x,st.y,now.x,now.y);
            if (ss==-1) continue;
            now.d=st.d;
            now.t=st.t+ss-1;
            Q.push(now);
            continue;
        }
        if (num[st.x][st.y]+1>4)
        {
            cnt++;
            have[st.x][st.y]=false;
            out[st.x][st.y]=st.t;
            num[st.x][st.y]=0;
//            printf("***%d\n",st.t);
            for (int i=0;i<4;i++)
            {
                int dx=st.x+dir[i][0];
                int dy=st.y+dir[i][1];
                if (!isok(dx,dy)) continue;
                int ss=Search(i,dx,dy,now.x,now.y);
                if (ss==-1)  continue;
                now.t=st.t+ss;
                now.d=i;
//                printf("%d %d %d++\n",now.x,now.y,now.t);
                if (now.t>T) continue;
                Q.push(now);
            }
        }
        else
            num[st.x][st.y]++;
    }
}

int main()
{
    int i,j,x,y,z;
    while (~scanf("%d%d%d%d",&r,&c,&n,&T))
    {
        memset(num,0,sizeof(num));
        memset(id,-1,sizeof(id));
        memset(out,0,sizeof(out));
        memset(have,false,sizeof(have));
        for (i=0;i<n;i++)
        {
            scanf("%d%d%d",&x,&y,&z);
            x--;
            y--;
            num[x][y]=z;
            have[x][y]=true;
            id[x][y]=i;
        }
        scanf("%d%d",&x,&y);
        x--;y--;
        have[x][y]=true;
        num[x][y]=10;
        bfs(x,y);
        for (i=0;i<r;i++)
        {
            for (j=0;j<c;j++)
            {
                if (id[i][j]==-1) continue;
                if (have[i][j])
                    q[id[i][j]][0]=1,q[id[i][j]][1]=num[i][j];
                else
                    q[id[i][j]][0]=0,q[id[i][j]][1]=out[i][j];

            }
        }
        for (i=0;i<n;i++)
        {
            printf("%d %d\n",q[i][0],q[i][1]);
        }
    }
    return 0;
}
/*
4 5 4 6
2 2 4
2 4 2
4 4 4
2 5 4
4 2
*/