UVALive - 3027
Description
A very big corporation is developing its corporative network. In the beginning each of the N enterprises of the corporation, numerated from 1 to N, organized its own computing and telecommunication center. Soon, Input Your program has to be ready to solve more than one test case. The first line of the input file will contains only the number T of the test cases. Each test will start with the number N of enterprises (5≤N≤20000). Then some number of lines (no more than 200000)
The test case finishes with a line containing the word O. The I commands are less than N. Output The output should contain as many lines as the number of E commands in all test cases with a single number each ? the asked sum of length of lines connecting the corresponding enterprise with its serving center. Sample Input 1 4 E 3 I 3 1 E 3 I 1 2 E 3 I 2 4 E 3 O Sample Output 0 2 3 5 Source |
昨天不知道怎么搞的，，网站崩溃了一天。。
并查集！！
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm> //用于使用abs
using namespace std;
const int maxn = 20000 + 10;
int pa[maxn], d[maxn];
int find(int x)
{ //路径压缩，同时维护d[i]：结点i到树根的距离
if(pa[x] != x)
{
int root = find(pa[x]);
d[x] += d[pa[x]];
return pa[x] = root;
}
else return x;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
int n, u, v;
char cmd[9];
scanf("%d", &n);
for(int i=0; i <= n; i++) { pa[i] = i; d[i] = 0; } //初始化，每个结点单独是一棵树
while(scanf("%s", cmd) && cmd[0] != 'O')
{
if(cmd[0] == 'E')
{
scanf("%d", &u);
find(u); //调用find函数，其实是维护d[i]的值
printf("%d\n", d[u]);
}
if(cmd[0] == 'I')
{
scanf("%d %d", &u, &v);
pa[u] = v; d[u] = abs(u - v) % 1000;
}
}
}
return 0;
}