pat 1034. Head of a Gang (30)

最新推荐文章于 2025-04-08 00:38:28 发布

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本文介绍了一种使用深度优先搜索(DFS)算法来寻找图中的连通分量的方法，并详细展示了如何通过DFS遍历图来确定每个连通分量的成员节点及最小代价。此外，还提供了一个完整的C++实现示例。

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DFS寻找连通分量

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<map>
#include<algorithm>
#define SIZE 2000
#define INF 0x7fffffff
using namespace std;

int graph[SIZE][SIZE], call[SIZE];
bool flag[SIZE];
map<string, int> hh;
map<int, string> rhh;
struct gang{
    vector<int>mem;
    string head;
    int min,fl;
}gg[SIZE];
void Search(int num, int n,int s){
    for (int i = 0; i < n;i++)
        if (!flag[i] && graph[s][i]){
            flag[i] = true;
            Search(num, n, i);
            gg[num].mem.push_back(i);
            gg[num].min += call[i];
        }
}
int DFS(int n){
    int num = 0;
    for (int i = 0; i < n;i++)
        if (!flag[i]){
            flag[i] = true;
            Search(num, n,i);
            gg[num].mem.push_back(i);
            gg[num].min += call[i];
            num++;
        }
    return num;
}

bool cmp(const struct gang &a, const struct gang &b){
    if (a.fl == b.fl)
        return a.head < b.head;
    else return a.fl < b.fl;
}
int main(){
    freopen("1.in", "r", stdin);
    int n, limit;
    scanf("%d%d", &n, &limit);
    int i,min,ele;
    char from[4], to[4];
    for (i = 0,ele = 0; i < n; i++){
        scanf("%s%s%d", from, to,&min);
        if (hh.find(from) == hh.end())
            hh[from] = ele++;
        if (hh.find(to) == hh.end())
            hh[to] = ele++;
        int s = hh[from], d = hh[to];
        graph[s][d] = graph[d][s] = min;
        call[s] += min;
        call[d] += min;
        if (rhh.find(s) == rhh.end())
            rhh[s] = from;
        if (rhh.find(d)==rhh.end())
            rhh[d] = to;
    }
    int numofgang = DFS(ele);
    int gang = 0;
    map<string, int>::iterator k;
    for (i = 0; i < numofgang; i++){
        if (gg[i].mem.size() < 3||gg[i].min/2<=limit){
            gg[i].fl = INF;
            continue;
        }
        int max = 0,index = -1;
        for (int j = 0; j < gg[i].mem.size();j++)
            if (call[gg[i].mem[j]] > max){
                max = call[gg[i].mem[j]];
                index = gg[i].mem[j];
            }
        gg[i].head = rhh[index];
        gang++;
    }
    sort(gg, gg + numofgang, cmp);
    printf("%d\n", gang);
    for (i = 0; i < gang; i++)
        cout << gg[i].head << ' ' << gg[i].mem.size() << endl;
    return 0;
}