1034 Head of a Gang

One way that the police finds the head of a gang is to check people’s phone calls. If there is a phone call between A and B, we say that A and B is related. The weight of a relation is defined to be the total time length of all the phone calls made between the two persons. A “Gang” is a cluster of more than 2 persons who are related to each other with total relation weight being greater than a given threshold K. In each gang, the one with maximum total weight is the head. Now given a list of phone calls, you are supposed to find the gangs and the heads.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive numbers N and K (both less than or equal to 1000), the number of phone calls and the weight threthold, respectively. Then N lines follow, each in the following format:

Name1 Name2 Time

where Name1 and Name2 are the names of people at the two ends of the call, and Time is the length of the call. A name is a string of three capital letters chosen from A-Z. A time length is a positive integer which is no more than 1000 minutes.

Output Specification:

For each test case, first print in a line the total number of gangs. Then for each gang, print in a line the name of the head and the total number of the members. It is guaranteed that the head is unique for each gang. The output must be sorted according to the alphabetical order of the names of the heads.

Sample Input 1:

8 59

AAA BBB 10

BBB AAA 20

AAA CCC 40

DDD EEE 5

EEE DDD 70

FFF GGG 30

GGG HHH 20

HHH FFF 10

Sample Output 1:

2

AAA 3

GGG 3

Sample Input 2:

8 70

AAA BBB 10

BBB AAA 20

AAA CCC 40

DDD EEE 5

EEE DDD 70

FFF GGG 30

GGG HHH 20

HHH FFF 10

Sample Output 2:

0

解题思路：一道并查集的题目，名字用hash的办法变成int类型，所以开个26*26*26的s数组就行。用一个数组记录每个人总的通话时间，用于最后找出head。并查集找出团队，再在团队中找到通话时间最长的人就是结果。因为结果需要按字母序输出，所以我先将结果保存下来，排序之后再输出的。

做题的时候并查集的压缩路径的地方写错了，导致测试点2 4一直答案错误，还是太粗心。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
int group[26 * 26 * 26];
int ti[26 * 26 * 26];
vector<int> groupCount[26 * 26 * 26];
struct res{
int name;
int size;
};
bool cmp(const res&r1, const res&r2){
return r1.name < r2.name;
}
int find(int x){
int res = x;
while (group[res] != res){
res = group[res];
}
while (x != res){
int temp = group[x];
group[x] = res;
x = temp;
}
return res;
}
void join(int x, int y){
int fx = find(x);
int fy = find(y);
if (fx != fy){
group[fy] = fx;
}
}
int main(){
for (int n, k; scanf("%d%d", &n, &k) != EOF;){
for (int i = 0; i < 26 * 26 * 26; i++){
group[i] = i;
ti[i] = 0;
groupCount[i] = vector<int>();
}
for (int i = 0; i < n; i++){
char p1[4];
char p2[4];
int temp;
scanf("%s %s %d", p1, p2, &temp);
int index1 = (p1[0] - 'A') * 26 * 26 + (p1[1] - 'A') * 26 + (p1[2] - 'A');
int index2 = (p2[0] - 'A') * 26 * 26 + (p2[1] - 'A') * 26 + (p2[2] - 'A');
join(index1, index2);
ti[index1] += temp;
ti[index2] += temp;
}
for (int i = 0; i < 26 * 26 * 26; i++){
find(i);
}
for (int i = 0; i < 26 * 26 * 26; i++){
if (ti[i]>0){
groupCount[group[i]].push_back(i);
}
}
vector<res>out;
for (int i = 0; i < 26 * 26 * 26; i++){
if (groupCount[i].size() > 2){
int name = groupCount[i][0];
int maxTime = ti[groupCount[i][0]];
int totalTime = 0;
for (int j = 0; j < groupCount[i].size(); j++){
totalTime += ti[groupCount[i][j]];
if (ti[groupCount[i][j]] >= maxTime){
maxTime = ti[groupCount[i][j]];
name = groupCount[i][j];
}
}
if (totalTime > k*2){
res temp;
temp.name = name;
temp.size = groupCount[i].size();
out.push_back(temp);
}
}
}
sort(out.begin(), out.end(), cmp);
printf("%d\n", out.size());
for (int i = 0; i < out.size(); i++){
printf("%c%c%c %d\n", (char)(out[i].name / (26 * 26) + 'A'), (char)((out[i].name % (26 * 26)) / 26 + 'A'), (char)(out[i].name % 26 + 'A'), out[i].size);
}
}
return 0;
}