题目粗心:
给定一个N个顶点和M条边的无向图,K个查问,每一个查问输出长度为n的门路,判断该门路是否是TS cycle或者TS simple cycle并输入题目要求的对应信息。
算法思路:
咱们应用邻接矩阵G保留该图的边权,而后应用path数组保留每一次查问的门路,判断该门路是否是TS cycle和TS simple cycle的办法如下:
- 1、应用 set differentVertices汇合保留所有门路上的不同顶点数目,total_dist为门路长度,frequencyOfStart为终点呈现的次数,isNa标记门路是否连通。
- 2、遍历门路上每一个顶点,累加total_dist的边权并判断以后边权是否为0,如果是,阐明该门路不连通,
isNa = true
,并统计终点呈现的次数。 - 3、首先判断isNa是否为true,如果是阐明以后门路不连通,肯定不是cycle,输入
printf("Path %d: NA (Not a TS cycle)n",i);
如果不是转4 - 4、判断终点和起点是否一样,如果不一样,阐明该门路不是cycle,输入
printf("Path %d: %d (Not a TS cycle)n",i,total_dist);
否则转5 - 5、判断differentVertices的大小(门路中呈现的不同顶点数目)是否等于N,如果不是,阐明没有拜访每一个城市,不是TS cycle或者TS simple cycle,输入
printf("Path %d: %d (Not a TS cycle)n",i,total_dist);
否则转6 - 6、判断顶点呈现了几次,如果是2次,阐明是TS simple cycle,输入
printf("Path %d: %d (TS simple cycle)n",i,total_dist);
否则是TS cycle,输入printf("Path %d: %d (TS cycle)n",i,total_dist);
同时在此状况下,得更新TS cycle或者TS simple cycle的最小门路长度和编号。
留神点:
- 1、只有在判断以后门路是TS cycle或者TS simple cycle的时候能力进行更新最小门路长度
提交后果:
AC代码:
#include<cstdio>#include<vector>#include<unordered_set>using namespace std;int N,M;// 顶点数目和边数int G[205][205];// 邻接矩阵,存储边权,不存在为0int main(){ scanf("%d %d",&N,&M); int a,b,dist; for (int i = 0; i < M; ++i) { scanf("%d %d %d", &a, &b, &dist); G[a][b] = G[b][a] = dist; } int K,num,city; scanf("%d",&K); int min_dis = 0x3fffffff; int min_dis_index = -1; for(int i=1;i<=K;++i){ scanf("%d",&num); vector<int> path; unordered_set<int> differentVertices;// 门路中不同顶点的数目 for (int j = 0; j < num; ++j) { scanf("%d",&city); path.push_back(city); } // 取得门路长度和终点呈现的次数 int total_dist = 0; int frequencyOfStart = 0; bool isNa = false; for (int k = 0; k < path.size(); ++k) { differentVertices.insert(path[k]); if(k<path.size()-1){ total_dist += G[path[k]][path[k+1]]; if(G[path[k]][path[k+1]]==0) { // 存在无奈达到的边 isNa = true; } } if(path[k]==path[0]){ ++frequencyOfStart; } } if(isNa){ // 以后门路不连通,肯定不是cycle printf("Path %d: NA (Not a TS cycle)\n",i); } else { // 肯定连通 if(path[0]!=path[path.size()-1]){ // 终点和起点不一样,不是cycle printf("Path %d: %d (Not a TS cycle)\n",i,total_dist); } else { // 肯定是cycle,终点至多呈现了2次 if(differentVertices.size()!=N){ // 没有拜访每一个城市,不是TS cycle或者TS simple cycle printf("Path %d: %d (Not a TS cycle)\n",i,total_dist); } else { // 肯定是TS cycle或者TS simple cycle if(min_dis>total_dist){ // 更新最短距离和编号 min_dis_index = i; min_dis = total_dist; } if(frequencyOfStart==2){ // 顶点只呈现了2次且拜访了每一个城市,是TS simple cycle printf("Path %d: %d (TS simple cycle)\n",i,total_dist); } else{ // 顶点呈现大于2次且拜访了每一个城市,是TS cycle printf("Path %d: %d (TS cycle)\n",i,total_dist); } } } } } printf("Shortest Dist(%d) = %d",min_dis_index,min_dis); return 0;}