uva 1599双向bfs

最新推荐文章于 2024-02-11 06:34:53 发布

原创最新推荐文章于 2024-02-11 06:34:53 发布 · 164 阅读

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本文介绍了一种使用双向BFS算法解决图中两点间最短路径问题的方法。通过正向和反向两次广度优先搜索，有效地减少了搜索范围，并在过程中实现了剪枝操作以提高效率。

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//0.140s
#include<iostream>
#include<queue>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 100005;
const int inf = 1 << 30;
int i, a, b, c, n, m;
bool vis[maxn];

struct node {
	int v; int c;
	node* next;
}G[maxn * 4];
node *adj[maxn], *cur, *p;
int path[maxn], d[maxn], w[maxn], bs;

void rev_bfs() {
	memset(d, 0, (n + 1) * sizeof(int));
	d[n] = 1; 
	queue<int> q;
	q.push(n);
	while (!q.empty()) {
		a = q.front();  q.pop();
		if (d[1] && d[a] >= d[1]) {
			continue;//剪枝1
		}
		for (p = adj[a]; p; p = p->next) {
			b = p->v;
			if (!d[b])
			{
				d[b] = d[a] + 1;
				q.push(b);
			}
		}
	}
}


void bfs() {
	memset(vis,0, sizeof(vis));
	vis[1] = true;
	vector<int> q1;
	q1.push_back(1);
	m = d[1]-1;
	while(m){
		bs = inf;
		for (i = 0; i < q1.size(); i++) {
			a = q1[i];
			for (p = adj[a]; p; p = p->next) {
				b = p->v; c = p->c;
				if (d[b] == m && c < bs)
					bs = c;
			}
		}
		vector<int> q2;
		for (i = 0; i < q1.size(); i++) {
			a = q1[i];
			for (p = adj[a]; p; p = p->next) {
				b = p->v;
				if (p->c == bs && d[b] == m && !vis[b])
				{
					vis[b] = true;//剪枝2
					q2.push_back(b); //cout<<a<<'-'<<bs<<'-' << p->v << endl;
				}
			}
		}
		w[m--] = bs;
		q1 = q2;
	}
	printf("%d\n", d[1]-1);
	printf("%d", w[d[1]-1]);
	for (i =d[1]-2; i >0; i--) {
		printf(" %d", w[i]);
	}
	printf("\n");
}


int main()
{
	while (scanf("%d%d", &n, &m) == 2) {
		memset(adj, 0, sizeof(adj));
		cur = G;
		for (i = 0; i < m; i++) {
			scanf("%d%d%d", &a, &b, &c);
			cur->v = b; cur->c = c; cur->next = adj[a];//从前面插入
			adj[a] = cur++;
			cur->v = a; cur->c = c; cur->next = adj[b];
			adj[b] = cur++;
		}
		rev_bfs();
		bfs();
	}
	return 0;
}

/*
4 6
1 2 1
1 3 2
3 4 3
2 3 1
2 4 4
3 1 1
*/