多校 hdu

时间：2015-07-30 23:26:18 阅读：348 评论：0 收藏：0 [点我收藏+]

标签：acm hdu

XYZ and Drops

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 430 Accepted Submission(s): 105

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 (

y), before the first second there is a waterdrop cracking at position (

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

n and

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

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

Sample Output

Author

XJZX

Source

2015 Multi-University Training Contest 4

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define maxn 100 + 5
struct Water
{
    int x, y, d;
}W[maxn*maxn], flag[maxn];
int time, r, c, n;
int d[maxn][maxn];
int go[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
int tt[maxn][maxn];
int main()
{
    while(~scanf("%d%d%d%d", &r, &c, &n, &time))
    {
        memset(d, 0, sizeof(d));
        memset(tt, -1, sizeof(tt));
        int x, y, num;
        queue<Water> Q;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d%d", &flag[i].x, &flag[i].y, &num);
            d[flag[i].x][flag[i].y] = num;
        }
        scanf("%d%d", &x, &y);
        for(int i=0; i<4; i++)
        Q.push(Water{x, y, i});
        int tmp, ti = 0;
        while(time--)
        {
            ti++;
            tmp = Q.size();
            if(!tmp)
            break;
            while(tmp--)
            {
                Water w = Q.front();
                Q.pop();
                int xx = w.x + go[w.d][0];
                int yy = w.y + go[w.d][1];
                if(xx > 0 && xx <= r && yy >=0 && yy <= c)
                if(d[xx][yy] == 0)
                {
                    Q.push(Water{xx, yy, w.d});
                }
                else d[xx][yy]++;
            }

            for(int i=1; i<=n; i++)
            {
                 x = flag[i].x, y = flag[i].y;
                 if(tt[x][y] == -1 && d[x][y] > 4)
                 {
                     d[x][y] = 0;
                     tt[x][y] = ti;
                     for(int i=0; i<4; i++)
                     Q.push(Water{x, y, i});
                 }
            }
        }

        for(int i=1; i<=n; i++)
        {
            x = flag[i].x, y = flag[i].y;
            if(tt[x][y] != -1)
                printf("0 %d\n", tt[x][y]);
            else
                printf("1 %d\n", d[x][y]);
        }
    }
    return 0;
}

多校 hdu

标签：acm hdu

原文地址：http://blog.csdn.net/dojintian/article/details/47157089

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