poj 2236 Wireless Network 【并查集】
Time Limit: 10000MS | Memory Limit: 65536K | |
Total Submissions: 16832 | Accepted: 7068 |
Description
one, and the network gradually began to work again. Because of the hardware restricts, each computer can only directly communicate with the computers that are not farther than d meters from it. But every computer can be regarded as the intermediary of the
communication between two other computers, that is to say computer A and computer B can communicate if computer A and computer B can communicate directly or there is a computer C that can communicate with both A and B.
In the process of repairing the network, workers can take two kinds of operations at every moment, repairing a computer, or testing if two computers can communicate. Your job is to answer all the testing operations.
Input
xi, yi (0 <= xi, yi <= 10000), which is the coordinate of N computers. From the (N+1)-th line to the end of input, there are operations, which are carried out one by one. Each line contains an operation in one of following two formats:
1. "O p" (1 <= p <= N), which means repairing computer p.
2. "S p q" (1 <= p, q <= N), which means testing whether computer p and q can communicate.
The input will not exceed 300000 lines.
Output
Sample Input
4 1
0 1
0 2
0 3
0 4
O 1
O 2
O 4
S 1 4
O 3
S 1 4
Sample Output
FAIL
SUCCESS
注意要区分修过的和没修过的
代码:
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define M 1005
#define LL __int64 struct node{
int x, y;
}s[M];
int map[M][M], fat[M], vis[M]; int f(int x){
if(fat[x] != x) fat[x] = f(fat[x]);
return fat[x];
} int main(){
int n, d;
scanf("%d%d", &n, &d);
int i, j;
d*=d;
for(i = 1; i <= n; i ++){
scanf("%d%d", &s[i].x, &s[i].y);
map[i][i] = 0;
for(j = i-1; j> 0; j --){
map[j][i] = map[i][j] = (s[i].x-s[j].x)*(s[i].x-s[j].x)+(s[i].y-s[j].y)*(s[i].y-s[j].y);
// printf("%d..map(%d, %d)\n", map[i][j], i, j);
}
}
char ss[2];
int a, b;
memset(vis, 0, sizeof(vis));
for(i = 1; i <= n; i++) fat[i] = i;
while(scanf("%s", ss) == 1){
if(ss[0] == 'O'){
scanf("%d", &a);
vis[a] = 1;
for(i = 1; i <= n; i ++){
if(vis[i]&&map[a][i] <= d&&i != a){
int x = f(a); int y = f(i);
if(x != y) fat[x] = y;
}
}
}
else{
scanf("%d%d", &a, &b);
int x = f(a); int y = f(b);
if(vis[a]&&vis[b]&&x == y) printf("SUCCESS\n");
else printf("FAIL\n");
}
}
return 0;
}
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