1034 Head of a Gang (30 point(s))
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
题目大意：
输入 N 条通话记录 Name1 Name2 和 Time，和一个值 K，有通话记录的人之间看作一个团伙，当团伙人数超过 2 人且通话总权值超过 K，则团伙中权值最大的人为头目；
输出满足条件的头目，和头目所在团伙的总人数，头目按字典排序输出；
设计思路：
DFS、连通分量
-
因为名字给的都是字母，所以在 C 中，还要考虑如何用数字保存和映射名字：
三个字母组成的名字，直接按照 26 进制处理三位数，转换成数字，最大值为 17575；用 id 和 name 两个数组分别记录每个人的序号和名字，序号按照每个人的出现次序依次递增，最终还获得了输入的总人数 count，
例如：ABC->28->id[28] = count->name[count] = ABC->count++ -
用数组储存边和节点的权值，DFS 遍历图，记录下符合条件的团伙
-
团伙输出前，依据头目名排序
编译器：C (gcc)
#include <stdio.h>
#include <string.h>
struct gang {
int num;
int member;
};
int toid(char s[]);
int dfs(int u, int *head, int *m, int *totalweight);
int cmp(const void *a, const void *b);
int g[2010][2010], w[2010], vis[2010];
int id[20000], count = 0;
char name[20000][4];
struct gang gangs[20000];
int sum = 0;
int n, k;
int main(void)
{
char s1[4] = {0}, s2[4] = {0};
int n1, n2, t;
int i;
scanf("%d %d", &n, &k);
for (i = 0; i < n; i++) {
scanf("%s %s %d", s1, s2, &t);
n1 = toid(s1);
n2 = toid(s2);
g[n1][n2] += t;
g[n2][n1] += t;
w[n1] += t;
w[n2] += t;
}
for (i = 1; i <= count; i++) {
if (vis[i] == 0) {
int head = i, m = 0, totalweight = 0;
dfs(i, &head, &m, &totalweight);
if (m > 2 && totalweight > k) {
gangs[sum].num = head;
gangs[sum].member = m;
sum++;
}
}
}
qsort(gangs, sum, sizeof(gangs[0]), cmp);
printf("%d\n", sum);
for (i = 0; i < sum; i++)
printf("%s %d\n", name[gangs[i].num], gangs[i].member);
return 0;
}
int cmp(const void *a, const void *b)
{
struct gang *s1 = (struct gang *)a, *s2 = (struct gang *)b;
int n1 = s1->num, n2 = s2->num;
return (strcmp(name[n1], name[n2]));
}
int dfs(int u, int *head, int *m, int *totalweight)
{
int v;
vis[u] = 1;
(*m)++;
if (w[u] > w[*head])
*head = u;
for (v = 1; v <= count; v++) {
if (g[u][v] > 0) {
(*totalweight) += g[u][v];
g[u][v] = g[v][u] = 0;
if (vis[v] == 0)
dfs(v, head, m, totalweight);
}
}
return 0;
}
int toid(char s[])
{
int n = 0;
int i = 0, temp = 0;
if (s[i] != '\0') {
n = (s[i] - 'A');
i++;
temp = 1;
}
while (s[i] != '\0') {
temp *= 26;
n += ((s[i] - 'A') * temp);
i++;
}
if (id[n] == 0) {
count++;
id[n] = count;
strcpy(name[count], s);
}
return id[n];
}