【图论05】并查集 1002 小希的迷宫

来源:互联网 发布:淘宝店主寄语 文艺 编辑:程序博客网 时间:2024/06/05 15:08

算法思路;1.经典的并查集;2.连通且不存在回路条件:边数 + 1 = 顶点数

注意陷阱(测试数据本身设计不太合理):

  0 0 Yes
  1 1 0 0  Yes


//模板开始#include <string>   #include <vector>   #include <algorithm>   #include <iostream>   #include <sstream>   #include <fstream>   #include <map>   #include <set>   #include <cstdio>   #include <cmath>   #include <cstdlib>   #include <ctime>#include<iomanip>#include<string.h>#define SZ(x) (int(x.size()))using namespace std;int toInt(string s){istringstream sin(s); int t; sin>>t; return t;}template<class T> string toString(T x){ostringstream sout; sout<<x; return sout.str();}typedef long long int64;int64 toInt64(string s){istringstream sin(s); int64 t; sin>>t;return t;}template<class T> T gcd(T a, T b){ if(a<0) return gcd(-a, b);if(b<0) return gcd(a, -b);return (b == 0)? a : gcd(b, a % b);}//模板结束(通用部分)#define ifs cinint findset(int x, int pa[]){return pa[x] != x ? pa[x] = findset(pa[x], pa) : x;}//【图论05】并查集 1002 小希的迷宫#define MAX_SIZE 100005int next_node[MAX_SIZE];//存储有向图的边//int in[MAX_SIZE];//存储节点的入度//int out[MAX_SIZE];//存储节点的出度int flag[MAX_SIZE];//标记节点是否存在void init()//初始化{for(int i = 0; i < MAX_SIZE; i++){next_node[i] = i;}//memset(in, 0, sizeof(in));//memset(out, 0, sizeof(out));memset(flag, 0, sizeof(flag));}int findset(int a)//找元素所在集合的代表元(因为用了路径压缩,路径压缩的主要目的是为了尽快的确定元素所在的集合){while(next_node[a] != a){a = next_node[a];}return a;}void union_nodes(int a, int b)//集合合并{int a1 = findset(a);int b1 = findset(b);next_node[a1] = b1;}int main(){//ifstream ifs("shuju.txt", ios::in);int m, n;while(ifs>>m>>n && !(m == -1 && n == -1)){if(m == 0 && n == 0){cout<<"Yes"<<endl;continue;}int bianshu = 0;init();bianshu++;union_nodes(m, n);flag[m]++;flag[n]++;while(ifs>>m>>n && !(m == 0 && n == 0))//输入数据,建立有向图,并合并相关集合{//int a = data[0] - 'a';//int b = data[strlen(data) - 1] - 'a';bianshu++;union_nodes(m, n);//out[a]++;//in[b]++;flag[m]++;flag[n]++;}int count = 0;for(int j = 0; j < MAX_SIZE; j++)//计算有向图中连通分支的个数{if(next_node[j] == j && flag[j] != 0){count++;}}int jiedianshu = 0;for(int j = 0; j < MAX_SIZE; j++){if(flag[j] != 0){jiedianshu++;}}if(jiedianshu == 1){cout<<"No"<<endl;}if(count == 1 && jiedianshu == bianshu + 1){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}}return 0;}