NCPC 2016 October 8,2016 Artwork
Problem A
Artwork
Problem ID: artwork
Time limit: 4 seconds
A template for an artwork is a white grid of n×m squares. The artwork will be created by painting q horizontal and vertical black strokes. A stroke starts from square (x1,y1), ends at square (x2,y2) (x1 = x2 or y1 = y2) and changes the color of all squares (x,y) to black where x1 ≤ x ≤ x2 and y1 ≤ y ≤ y2. The beauty of an artwork is the number of regions in the grid. Each region consists of one or more white squares that are connected to each other using a path of white squares in the grid, walking horizontally or vertically but not diagonally. The initial beauty of the artwork is 1. Your task is to calculate the beauty after each new stroke. Figure A.1 illustrates how the beauty of the artwork varies in Sample Input 1.
1 region 3 regions 3 regions 4 regions 3 regions
Figure A.1: Illustration of Sample Input 1.
Input The first line of input contains three integers n, m and q (1 ≤ n,m ≤ 1000, 1 ≤ q ≤ 104). Then follow q lines that describe the strokes. Each line consists of four integers x1, y1, x2 and y2 (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ m). Either x1 = x2 or y1 = y2 (or both).
Input
The first line of input contains three integers n, m and q (1 ≤ n,m ≤ 1000, 1 ≤ q ≤ 104). Then follow q lines that describe the strokes. Each line consists of four integers x1, y1, x2 and y2 (1 ≤ x1 ≤ x2 ≤ n, 1 ≤ y1 ≤ y2 ≤ m). Either x1 = x2 or y1 = y2 (or both).
Output
For each of the q strokes, output a line containing the beauty of the artwork after the stroke.
Sample Input 1 Sample Output 1
4 6 5 2 2 2 6 1 3 4 3 2 5 3 5 4 6 4 6 1 6 4 6 | 1 3 3 4 3 |
这题输出方式让我想死。。。之前训练比赛时认为是输入一行就输出一行。。结果一直output limit exceeded。。后面看了别人的才发现是输完后再一起输出。。想死的心都有了。
思路：从最后一张图开始往前推，再用并查集解决。详细看代码
#include<stdio.h>
#include<string.h>
#define MAXN 1010
struct stock{
int x1, x2, y1, y2;
}pos[10010];
int num[MAXN*MAXN], fa[MAXN*MAXN],ans[10010];//num表示当前位置涂了几个黑点 ans[n]表示第n次画横线时白色连通块的个数
int dir[4][2] = { 0, 1, -1, 0, 0, -1, 1, 0 };
int m, n, p,t;
int hash(int x, int y)//用哈希表表示
{
int num = (x - 1)*m + y;
return num;
}
void init()//初始化
{
for (int i = 1; i <= n*m; i++)
{
fa[i] = i;
num[i] = 0;
}
}
int find(int x)
{
return x == fa[x] ? x : fa[x] = find(fa[x]);
}
void merge(int x, int y)//将两个相邻连通块连接在一起
{
int fx = find(x), fy = find(y);
if (fx == fy)
return;
t--;//t表示连通块个数
fa[fx] = fy;
}
bool check(int x, int y)
{
if (x >= 1 && y >= 1 && x <= n&&y <= m)
return true;
return false;
}
void work(int x,int y)//将刚出现的白块连到连通块中
{
for (int i = 0; i < 4; i++)
{
int xx = x + dir[i][0];
int yy = y + dir[i][1];
if (check(xx, yy) && !num[hash(xx,yy)])
{
merge(hash(xx,yy), hash(x,y));
}
}
}
int main()
{
scanf("%d %d %d", &n, &m, &p);
t = m*n;
init();
//画黑线
for (int i = 1; i <= p;i++)
{
scanf("%d%d%d%d", &pos[i].x1, &pos[i].y1, &pos[i].x2, &pos[i].y2);
for (int x = pos[i].x1; x <= pos[i].x2; x++)
{
for (int y = pos[i].y1; y <= pos[i].y2; y++)
{
if (num[hash(x, y)] == 0)
t--;
num[hash(x, y)]++;
}
}
}
//求出最后一个图的白色连通块的个数
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
if (!num[hash(i,j)])
{
work(i, j);
}
}
}
//向前面的图推
for (int i = p; i > 0; i--)
{
ans[i] = t;
//一步一步撤去黑线
for (int x = pos[i].x1; x <= pos[i].x2; x++)
{
for (int y = pos[i].y1; y <= pos[i].y2; y++)
{
num[hash(x, y)]--;
if (num[hash(x, y)] == 0)
{
t++;//黑块撤完白块数目增加
work(x, y);
}
}
}
}
for (int i = 1; i <= p; i++)
{
printf("%d\n", ans[i]);
}
return 0;
}在比赛时从正面进行时是通过黑块入手的，发现时间没有超，但一直output limit exceeded
