百度笔试914,运维开发第二题代码
#include <iostream>
#include<vector>
using namespace std;
int main() {
int output;
vector<vector<int> > res;
vector<vector<char> > contain;
int n, m;
int min, max;
cin >> n >> m;
char query;
for (int i = 0; i < m; i++) {
vector<char> con1;
for (int j = 0; j < 3; j++) {
cin >> query;
con1.push_back(query);
}
contain.push_back(con1);
}
for (int i = 0; i < n; i++) {
vector<int> con;
int temp;
con.push_back(i + 1);
con.push_back(i+1);
res.push_back(con);
}
//截至到这里都是在存储输入数据,下面对输入进行处理;
//分两种情况,一个是发指令,一个是询问。
int a, b;
for (int i = 0; i < m; i++) {
if (contain[i][0] == 'C') {
a = int(contain[i][1]-'0');
b = int(contain[i][2]-'0');
for (int i = 1; i < res[a - 1].size(); i++) {
res[b - 1].push_back(res[a - 1][i]);
}
}
else if(contain[i][0] == 'Q') {
output = -1;
max = min = 0;
a = int(contain[i][1]-'0');
b = int(contain[i][2]-'0');
for (int j = 0; j < n; j++) {
if (res[j].size() != 2) {
for (int i = 1; i < res[j].size(); i++) {
if (res[j][i] == a) min = i;
else if (res[j][i] == b) max = i;
if (min != 0 && max != 0) output= abs(min - max)-1;
}
if (output != -1) {
cout << output << endl;
break;
}
}
if (j == n - 1) cout << -1 << endl;
}
}
}
return 0;
}
第一题暴力查找,时间超限制,求简便方法
全部评论
(0) 回帖