- 题目描述：
-
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 threthold 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.
- 输入：
-
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.
- 输出：
-
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.
- 样例输入：
-
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 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
- 样例输出：
-
2 AAA 3 GGG 3 0
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 #define N 1001 7 typedef struct node{ 8 char name[4];//名字 9 int time;//通话时间 10 }Node; 11 typedef struct gang{ 12 int head;//首领编号 13 // int mem[N];//成员的编号 14 int num;//成员数目 15 int weight;//总共通话时间 16 }Gang; 17 Node people[N];//用于记录每个人的信息 right 18 int Tree[N];//用于并查集 right 19 int k;//阈值 right 20 Gang group[N];//统计每个组的情况 right 21 int findRoot(int x){ 22 if(Tree[x]==-1) return x; 23 else{ 24 int tmp=findRoot(Tree[x]); 25 Tree[x]=tmp; 26 return tmp; 27 } 28 } 29 Gang* qsort(int Tree[],Gang group[],int num,int &count){//挑出符合的 30 int i; 31 //int count=0; 32 for(i=0;i<num;i++){ 33 if(Tree[i]==-1&&group[i].weight>k&&group[i].num>2) count++; 34 } 35 Gang *ans; 36 ans=(Gang*) malloc(sizeof(Gang)*count); 37 int j=0; 38 for(i=0;i<num;i++){ 39 if(Tree[i]==-1&&group[i].weight>k&&group[i].num>2){ 40 ans[j].head=group[i].head; 41 ans[j].num=group[i].num; 42 ans[j].weight=group[i].weight; 43 j++; 44 } 45 } 46 return ans; 47 48 } 49 bool cmp(Gang a,Gang b){ 50 51 return (strcmp(people[a.head].name,people[b.head].name)<0); 52 } 53 int main(){ 54 int n;//n 是记录数，k是阈值 55 while(scanf("%d%d",&n,&k)!=EOF){ 56 int num=0;//当前人数 57 int i; 58 for(i=0;i<N;i++) Tree[i]=-1; 59 //for(i=1;i<N;i++) sum[i]=0; 60 while(n--!=0){ 61 char name1[4],name2[4]; 62 int s1,s2;//记录name在数组中位置 63 int weight;//两人通话时长 64 scanf("%s%s%d",name1,name2,&weight); 65 for(i=0;i<num;i++){ 66 if(strcmp(name1,people[i].name)==0) break;//若已经存在 67 } 68 if(i==num){ 69 strcpy(people[i].name,name1); 70 people[i].time=0; 71 group[i].num=1; 72 group[i].weight=0; 73 group[i].head=i; 74 num++; 75 } 76 s1=i; 77 people[i].time+=weight; 78 79 for(i=0;i<num;i++){ 80 if(strcmp(name2,people[i].name)==0) break; 81 } 82 if(i==num){ 83 strcpy(people[i].name,name2); 84 people[i].time=0; 85 group[i].num=1; 86 group[i].weight=0; 87 group[i].head=i; 88 num++; 89 } 90 s2=i; 91 people[i].time+=weight; 92 93 int s1_=findRoot(s1); 94 int s2_=findRoot(s2); 95 if(people[s1].time>people[group[s1_].head].time){//更新组里面的head 96 group[s1_].head=s1; 97 } 98 if(people[s2].time>people[group[s2_].head].time){//更新组里面的head 99 group[s2_].head=s2; 100 } 101 if(s1_!=s2_) {//合并 102 Tree[s1_]=s2_; 103 group[s2_].num+=group[s1_].num; 104 group[s2_].weight+=(weight+group[s1_].weight); 105 if(people[group[s2_].head].time<people[group[s1_].head].time){ 106 group[s2_].head=group[s1_].head;// 107 } 108 109 } 110 else group[s2_].weight+=weight; 111 112 113 114 115 116 } 117 Gang* ans; 118 int count=0; 119 ans=qsort(Tree,group,num,count); 120 sort(ans,ans+count,cmp); 121 printf("%d\n",count); 122 for(i=0;i<count;i++){ 123 printf("%s %d\n",people[ans[i].head].name,ans[i].num); 124 } 125 126 } 127 128 return 0; 129 }
时间： 2024-10-10 01:29:58