2015 多校赛 第四场 1010 (hdu 5336)

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

 

题意：

在一个 r*c 的格子图上，有 n 个水滴静止。每当水滴体积超过 4 时，会分成 4 个体积为 1 的小水滴向上下左右四个方向运动，每秒移动一个格子。小水滴与静止的水滴碰撞后会合成，体积相加。开始时 n 个水滴都静止，给出一个在 (x,y) 位置炸开的水滴作为启动源。

注意：1.小水滴彼此间不影响，即使同时经过同一位置。

2.若多个小水滴同时到达同一格子，则都与位于该格子的水滴合体，体积相加。超过 4 后按上述规则分裂。（即使为6,7等等）

 

思路：

因为数据较小所以直接暴力模拟O(nT)也能过。。

笔者用优先队列维护最先受到影响的水滴，依次处理，直到时间超出 T 或者是优先队列为空，则终止。

刚开始没有注意到应该注意的两点各种跪。。

水滴的结构体中还有一个变量 dir 记录该影响移动的方向。因为存在到了该“影响”发生的时候原有位置的水滴已经分裂不存在的情况，此时应继续向该方向搜索离之最近的点。

感觉实现的时候还是比较多细节要注意的。

代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
#define mod 1005
const int U=1,D=2,L=3,R=4;
int r,c,n,t,x,y;
struct drop{
    int x,y,size,status,time,num,dir;
    bool operator < (const drop d) const{
        return time>d.time;
    }
}drops[105];
int update(int num,int dir){
    int closest=-1;
    for(int i=0;i<n;i++) if(drops[i].status&&i!=num){
        if(dir==U){
            if(drops[num].x==drops[i].x&&drops[num].y<drops[i].y)
                if(closest==-1||drops[i].y<drops[closest].y) closest=i;
        }
        else if(dir==D){
            if(drops[num].x==drops[i].x&&drops[num].y>drops[i].y)
                if(closest==-1||drops[i].y>drops[closest].y) closest=i;
        }
        else if(dir==L){
            if(drops[num].y==drops[i].y&&drops[num].x>drops[i].x)
                if(closest==-1||drops[i].x>drops[closest].x) closest=i;
        }
        else{
            if(drops[num].y==drops[i].y&&drops[num].x<drops[i].x)
                if(closest==-1||drops[i].x<drops[closest].x) closest=i;
        }
    }
    return closest;
}
priority_queue<drop>pq;
vector<drop> T;
int main(){
    for(int i=0;i<105;i++) drops[i].num=i;
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(~scanf("%d%d%d%d",&r,&c,&n,&t)){
        for(int i=0;i<n;i++){
            scanf("%d%d%d",&drops[i].x,&drops[i].y,&drops[i].size);
            drops[i].status=1,drops[i].time=10005;
        }
        scanf("%d%d",&x,&y);
        while(!pq.empty()) pq.pop();
        drops[100].x=x,drops[100].y=y,drops[100].status=1;
        drops[100].size=4,drops[100].time=0;
        pq.push(drops[100]);
        int t_now=0;
        while(!pq.empty()){
            drop tmp=pq.top();
            while(!drops[tmp.num].status){
                pq.pop();
                int num=update(tmp.num,tmp.dir);
                if(num!=-1){
                    drop temp=drops[num];
                    temp.dir=tmp.dir;
                    if(temp.dir==U) temp.time=temp.y-tmp.y;
                    else if(temp.dir==D) temp.time=tmp.y-temp.y;
                    else if(temp.dir==L) temp.time=tmp.x-temp.x;
                    else temp.time=temp.x-tmp.x;
                    temp.time+=tmp.time;
                    pq.push(temp);
                }
                if(pq.empty()) break;
                else tmp=pq.top();
            }
            if(pq.empty()) break;
            if(tmp.time>t) break;
            int up=-1,down=-1,left=-1,righ=-1;
            if(drops[tmp.num].size<4){
                drops[tmp.num].size++;
                pq.pop();
                continue;
            }
            pq.pop();//new
            T.clear();
            while(!pq.empty()){
                drop Tmp=pq.top();
                if(Tmp.time==tmp.time&&Tmp.x==tmp.x&&Tmp.y==tmp.y)
                    pq.pop();
                else if(Tmp.time==tmp.time){
                    T.push_back(Tmp);
                    pq.pop();
                }
                else break;
            }
            for(int i=0;i<T.size();i++) pq.push(T[i]);
            drops[tmp.num].status=0;
            drops[tmp.num].time=tmp.time;
            t_now=drops[tmp.num].time;
            for(int i=0;i<n;i++) if(drops[i].status){
                if(tmp.x==drops[i].x){
                    if(tmp.y<drops[i].y){
                        if(up==-1||drops[i].y<drops[up].y) up=i;
                    }
                    else{
                        if(down==-1||drops[i].y>drops[down].y) down=i;
                    }
                }
                else if(tmp.y==drops[i].y){
                    if(tmp.x>drops[i].x){
                        if(left==-1||drops[i].x>drops[left].x) left=i;
                    }
                    else{
                        if(righ==-1||drops[i].x<drops[righ].x) righ=i;
                    }
                }
            }
            if(up!=-1&&drops[up].time>drops[up].y-tmp.y+t_now){
                drop temp=drops[up];
                temp.time=drops[up].y-tmp.y+t_now,temp.dir=U;
                pq.push(temp);
            }
            if(down!=-1){
                drop temp=drops[down];
                temp.time=tmp.y-drops[down].y+t_now,temp.dir=D;
                pq.push(temp);
            }
            if(left!=-1){
                drop temp=drops[left];
                temp.time=tmp.x-drops[left].x+t_now,temp.dir=L;
                pq.push(temp);
            }
            if(righ!=-1){
                drop temp=drops[righ];
                temp.time=drops[righ].x-tmp.x+t_now,temp.dir=R;
                pq.push(temp);
            }
            //pq.pop();
        }
        for(int i=0;i<n;i++){
            if(drops[i].status) printf("1 %d\n",drops[i].size);
            else printf("0 %d\n",drops[i].time);
        }
    }
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/names-yc/p/4703137.html