BJFUOJ-285

基于邻接表的长度为k的简单路径的求解

#include<iostream>
using namespace std;
#define maxsize 100
typedef struct ArcNode {
    int val;
    ArcNode* next;
    ArcNode():val(0),next(nullptr){}
    ArcNode(int x):val(x),next(nullptr){}

};
typedef struct VexNode {
    char data;
    ArcNode* first;
    VexNode():data('0'),first(nullptr){}
}VexNode;

typedef struct {
    int VexNum;
    int ArcNum;
    VexNode vexs[maxsize];
}ALGraph;

int findindex(ALGraph G, char x) {
	for (int i = 1; i <= G.VexNum; i++) {
		if (G.vexs[i].data == x) return i;
	}
	return 0;
}
void CreateGraph(ALGraph& G, int n, int m) {
	G.VexNum = n;
	G.ArcNum = m;
	for (int i = 1; i <= n; i++) { //s输入顶点
		char x;
		cin >> x;
		G.vexs[i].data = x;
	}
	for (int i = 1; i <= m; i++) { //输入边
		char a1, a2;
		cin >> a1 >> a2;

		int index1 = findindex(G, a1);
		int index2 = findindex(G, a2);
		ArcNode* p1 = new ArcNode(index1);
		ArcNode* p2 = new ArcNode(index2);
		p1->next = G.vexs[index2].first;
		G.vexs[index2].first = p1;

		p2->next = G.vexs[index1].first;
		G.vexs[index1].first = p2;
	}

}
void dfs(ALGraph G, int start, int end, int n, int* visited, int& flag, int k) {

	if (start == end  && n == k) {
		flag = 1;
		return;
	}
	ArcNode* p = G.vexs[start].first;
	while (p) {
		if (visited[p->val] == 0) {
			visited[p->val] = 1;
			dfs(G, p->val, end, n + 1, visited, flag, k);
		}
		p = p->next;
	}
}
void findPathK(ALGraph G, char start,char end,int k) {
	int s = findindex(G,start);
	int e = findindex(G,end);
	int visited[30] = { 0 };
	int flag = 0;
	dfs(G, s, e,0, visited, flag, k);
	if (flag == 1) {
		cout << "YES" << endl;
	}
	else {
		cout << "NO" << endl;
	}


}


int main() {
	int n, m; int k;
	while (cin >> n >> m>>k && (n != 0 && m != 0 && k!=0)) {
		ALGraph G;
		CreateGraph(G, n, m);
		char v, d;
		cin >> v >> d;
		findPathK(G, v, d, k);
	}
}