hdu5336XYZ and Drops

题意：给出r*c的网格，有的网格为空。有的有水。再给出一个爆炸点，从这个点向四周爆出四个水滴，若碰到水则融为一体，若碰到其它水滴直接跑过去互不影响。每秒可跑一格，若水中水滴数量超过4则爆开。问T秒后网格的状态是如何的。

做法：因为数据有点多，直接用set优化bfs一次走一步的过程。变成一次走多步就可以。

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int num[110][110];
set<int>sr[110],sc[110];
int r,c,T;
bool beyond(int tr,int tc)
{
    return tr<1||tc<1||tr>r||tc>c;
}
struct point
{
    int tr,tc,flag;
    point(){}
    point(int tr,int tc,int flag)
    {
        this->tr=tr;
        this->tc=tc;
        this->flag=flag;
    }
}hehe[110];
vector<point>bx;
bool vis[110][110];
void gao(int tr,int tc,int flag,int cnt)
{
    if(!beyond(tr,tc)&&!vis[tr][tc])
    {
        if(num[tr][tc]>0)
        {
            num[tr][tc]++;
            if(num[tr][tc]>4)
            {
                sr[tr].erase(tc);
                sc[tc].erase(tr);
                num[tr][tc]=-cnt;
                bx.push_back(point(tr,tc,0));
                vis[tr][tc]=1;
            }
        }
        else
            bx.push_back(point(tr,tc,flag));
    }
}
void work()
{
    set<int>::iterator it;
    int cnt=0,p=0;
    while(1)
    {
        int len=bx.size(),mn=INT_MAX;
        if(p==len)
            return;
        for(int i=p;i<len;i++)
        {
            point t=bx[i];
            int tr=t.tr,tc=t.tc,flag=t.flag;
            if(flag==0||flag==1)
            {
                it=sc[tc].lower_bound(tr);
                if(it!=sc[tc].begin())
                {
                    it--;
                    mn=min(mn,tr-*it);
                }
            }
            if(flag==0||flag==2)
            {
                it=sc[tc].upper_bound(tr);
                if(it!=sc[tc].end())
                    mn=min(mn,*it-tr);
            }
            if(flag==0||flag==3)
            {
                it=sr[tr].lower_bound(tc);
                if(it!=sr[tr].begin())
                {
                    it--;
                    mn=min(mn,tc-*it);
                }
            }
            if(flag==0||flag==4)
            {
                it=sr[tr].upper_bound(tc);
                if(it!=sr[tr].end())
                    mn=min(mn,*it-tc);
            }
        }
        if(cnt+mn>T)
            return;
        cnt+=mn;
        memset(vis,0,sizeof(vis));
        for(int i=p;i<len;i++)
        {
            point t=bx[i];
            int tr=t.tr,tc=t.tc,flag=t.flag;
            if(flag==0||flag==1)
            {
                tr-=mn;
                gao(tr,tc,1,cnt);
                tr+=mn;
            }
            if(flag==0||flag==2)
            {
                tr+=mn;
                gao(tr,tc,2,cnt);
                tr-=mn;
            }
            if(flag==0||flag==3)
            {
                tc-=mn;
                gao(tr,tc,3,cnt);
                tc+=mn;
            }
            if(flag==0||flag==4)
            {
                tc+=mn;
                gao(tr,tc,4,cnt);
                tc-=mn;
            }
        }
        p=len;
    }
}
int main()
{
    int n;
    while(scanf("%d%d%d%d",&r,&c,&n,&T)!=EOF)
    {
        for(int i=1;i<=r;i++)
            sr[i].clear();
        for(int i=1;i<=c;i++)
            sc[i].clear();
        memset(num,0,sizeof(num));
        for(int i=0;i<n;i++)
        {
            int t;
            scanf("%d%d%d",&hehe[i].tr,&hehe[i].tc,&t);
            sr[hehe[i].tr].insert(hehe[i].tc);
            sc[hehe[i].tc].insert(hehe[i].tr);
            num[hehe[i].tr][hehe[i].tc]=t;
        }
        int tr,tc;
        scanf("%d%d",&tr,&tc);
        bx.clear();
        bx.push_back(point(tr,tc,0));
        work();
        for(int i=0;i<n;i++)
        {
            tr=hehe[i].tr;
            tc=hehe[i].tc;
            int t=num[tr][tc];
            if(t<0)
                printf("0 %d\n",-t);
            else
                printf("1 %d\n",t);
        }
    }
}

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

Total Submission(s): 414    Accepted Submission(s): 101

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