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  • Uva




    先倒着BFS,找到每个结点i到重点最短距离d[i],然后直接起点开始BFS,如果有多种走法,选颜色字典序最小的走。

    AC代码:

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cctype>
    #include <cstring>
    #include <string>
    #include <sstream>
    #include <vector>
    #include <set>
    #include <map>
    #include <algorithm>
    #include <stack>
    #include <queue>
    #include <bitset> 
    #include <cassert> 
    
    using namespace std;
    
    const int maxn = 100005;
    const int maxc = 1000000000;
    
    struct Edge {
    	int u, v, c;
    	Edge(int u = 0, int v = 0, int c = 0) : u(u), v(v), c(c) {}
    };
    
    vector<Edge> edges;
    vector<int> G[maxn];
    
    void addEdge(int u, int v, int c)
    {
    	edges.push_back(Edge(u, v, c));
    	int idx = edges.size() - 1;
    	G[u].push_back(idx);
    }
    
    int n, vis[maxn];
    int d[maxn]; // 每个结点i到重点的最短距离d[i]
    
    // 从终点倒着BFS,找到到达n-1的最短路
    void revBfs()
    {
    	memset(vis, 0, sizeof(vis));
    	d[n - 1] = 0;
    	vis[n - 1] = true; // 访问状态
    
    	queue<int> q;
    	q.push(n - 1);
    	while (!q.empty()) {
    		int v = q.front();
    		q.pop();
    		for (int i = 0; i < G[v].size(); i++) {
    			int e = G[v][i];
    			int u = edges[e].v;
    			if (!vis[u]) { // 还没有杯访问过
    				vis[u] = true;
    				d[u] = d[v] + 1;
    				q.push(u);
    			}
    		}
    	}
    }
    
    vector<int> ans;
    
    // 从起点开始BFS,如果有多种走法,选颜色字典序最小的走
    void bfs()
    {
    	memset(vis, 0, sizeof(vis));
    	vis[0] = true;
    	ans.clear();
    
    	vector<int> next;
    	next.push_back(0);
    	for (int i = 0; i < d[0]; i++) {
    		int minColor = maxc;
    		for (int j = 0; j < next.size(); j++) {
    			int u = next[j];
    			for (int k = 0; k < G[u].size(); k++) {
    				int e = G[u][k];
    				int v = edges[e].v;
    				if (d[u] == d[v] + 1) {
    					minColor = min(minColor, edges[e].c);
    				}
    			}
    		}
    		ans.push_back(minColor);
    		
    		// 再一次BFS
    		vector<int> next2;
    		for (int j = 0; j < next.size(); j++) {
    			int u = next[j];
    			for (int k = 0; k < G[u].size(); k++) {
    				int e = G[u][k];
    				int v = edges[e].v;
    				if (d[u] == d[v] + 1 &&
    					!vis[v] &&
    					edges[e].c == minColor) {
    					vis[v] = true;
    					next2.push_back(v);
    				}
    			}
    		}
    		next = next2;
    	}
    
    	printf("%d
    ", ans.size());
    	printf("%d", ans[0]);
    	for (int i = 1; i < ans.size(); i++) {
    		printf(" %d", ans[i]);
    	}
    	printf("
    ");
    }
    
    int main()
    {
    	int u, v, c, m;
    	while (scanf("%d%d", &n, &m) == 2) {
    		edges.size();
    		for (int i = 0; i < n; i++) {
    			G[i].clear();
    		}
    		while (m--) {
    			scanf("%d%d%d", &u, &v, &c);
    			addEdge(u - 1, v - 1, c);
    			addEdge(v - 1, u - 1, c);
    		}
    		revBfs();
    		bfs();
    	}
    
    	return 0;
    }




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  • 原文地址:https://www.cnblogs.com/zhangyaoqi/p/4591547.html
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