1034. Head of a Gang (30)

题目连接：https://www.patest.cn/contests/pat-a-practise/1034

这道题我自己写的代码中没用到什么数据结构，就是普通的逻辑关系，求和，比大小，但有好3个测试点没过：

#include<stdio.h>
#include<stdlib.h>
#include<vector>
#include<string.h>

using namespace std;
typedef struct PNode{
    char ID[4];
    int time;
}per;

int N,K;
vector<per>Gang[3005];

int Have(char *s,vector<per>&p)
{
    int i,flag=-1;
    for (i=0;i<p.size();i++)
    {
        if (strcmp(s,p[i].ID)==0)
        {
            flag=i;
            break;
        }
    }
    return flag;
}

int cmp(const void *a,const void *b)
{
    return(strcmp((*(per*)a).ID,(*(per*)b).ID)>0);
}

int main()
{
    scanf("%d %d",&N,&K);
    char A1[4],A2[4];
    int t,i,index=0,j,t1,t2;
    per tmp1,tmp2;
    int weight[3005]={0};

    tmp1.time=tmp2.time=0;
    scanf("%s%*c%s%d",&tmp1.ID,&tmp2.ID,&t);
    tmp1.time+=t;
    tmp2.time+=t;
    Gang[0].push_back(tmp1);
    Gang[0].push_back(tmp2);
    weight[0]=t;
   // printf("%s
",tmp2.ID);
    int fg=0;
    getchar();
    for (i=1;i<N;i++)
    {
        scanf("%s%*c%s%d",&tmp1.ID,&tmp2.ID,&t);
        getchar(); //printf("---%s %s
",tmp1.ID,tmp2.ID);
        fg=0;
        tmp1.time=tmp2.time=t;
        for (j=0;j<=index;j++)
        {
            t1=Have(tmp1.ID,Gang[j]);
            t2=Have(tmp2.ID,Gang[j]);
            if (t1>=0 && t2 <0)
            {Gang[j][t1].time+=t;
            Gang[j].push_back(tmp2);
            fg=1;
            weight[j]+=t;
            }
            else if (t2>=0 &&t1<0){
                Gang[j][t2].time+=t;
                Gang[j].push_back(tmp1);
                fg=1;
                weight[j]+=t;
            }
            else if (t1>=0 && t2>=0)
            {
                Gang[j][t1].time+=t;
                Gang[j][t2].time+=t;
                fg=1;
                weight[j]+=t;
            }
        }
        if (fg==0)
        {
            index++;
            Gang[index].push_back(tmp1);
            Gang[index].push_back(tmp2);
            weight[index]=t;
        }
    }
   // printf("%d
",index);
    per Head[3005];
    int cnt=0,tmpj,Maxt=-1;

    for (i=0;i<=index;i++)
    {
       // printf("%d %d
",Gang[i].size(),weight[i]);
        if (Gang[i].size()>2 &&weight[i]>K)
        {
            for (j=0;j<Gang[i].size();j++)
            {
                if (Gang[i][j].time>Maxt)
                {
                    Maxt=Gang[index][j].time;
                    tmpj=j;
                }
            }
            strcpy(Head[cnt].ID,Gang[i][tmpj].ID);
            Head[cnt].time=Gang[i].size();
            cnt++;
        }
    }

    if (cnt==0){printf("0");return 0;}
    qsort(Head,cnt,sizeof(Head[0]),cmp);
    printf("%d
",cnt);
    for (i=0;i<cnt;i++)
    {
        printf("%s %d
",Head[i].ID,Head[i].time);
    }

    return 0;
}

希望大神能指出错误或提出测试样例。

参考了其他人的写法，才知道用到了映射，并查集/DFS等，主要参考了http://blog.csdn.net/xianyun2009/article/details/50176911的代码：

#include<stdio.h>
#include<vector>
#include<map>
#include<string>

using namespace std;

struct NODE{
    char a[4],b[4];
    int v;
}node[1011];

struct RES{
    string maxName;
    int maxV,totalV,member;
    RES():maxV(0),totalV(0),member(0){}
};

map<string,int>weight,lst;
int index[3000]={0};

int Find(int a)
{
    if (index[a]<0)return a;
    else return index[a]=Find(index[a]);
}

void Union(int a,int b)
{
    int a1,b1;
    a1=Find(a);
    b1=Find(b);
    if (a1!=b1)
    {
        if (index[a1]<=index[b1])
        {
            index[a1]+=index[b1];
            index[b1]=a1;
        }
        else if (index[a1]>index[b1])
        {
            index[b1]+=index[a1];
            index[a1]=b1;
        }
    }
}

int main()
{
    int N,K;
    scanf("%d %d",&N,&K);
    int i;
   // char namea[4],nameb[4];
    for (i=0;i<N;i++)
    {
        scanf("%s%s%d",&node[i].a,&node[i].b,&node[i].v);
        weight[node[i].a]+=node[i].v;
        weight[node[i].b]+=node[i].v;
    }

    int cnt=0;
    for (map<string,int>::iterator it =weight.begin();it!=weight.end();it++)
    {
        index[cnt]=-1;
        lst[it->first]=cnt++;
    }

    for (i=0;i<N;i++)
    {
        Union(lst[node[i].a],lst[node[i].b]);
    }

    map<int,RES>res;

    for (map<string,int>::iterator it=weight.begin();it!=weight.end();it++)
    {
        int fa=Find(lst[it->first]);
        if (it->second>res[fa].maxV)
        {
            res[fa].maxV=it->second;
            res[fa].maxName=it->first;
        }
        res[fa].totalV+=it->second;
        res[fa].member++;
    }

    map<string,int>rs;

    for (map<int,RES>::iterator it=res.begin();it!=res.end();it++)
    {
        if (((it->second).member)>2 && ((it->second).totalV)/2>K)
        {
            rs[(it->second).maxName]=(it->second).member;
        }
    }

    printf("%d
",rs.size());

    for (map<string,int>::iterator it=rs.begin();it!=rs.end();it++)
    {
        printf("%s %d
",it->first.c_str(),it->second);
    }

    return 0;
}