UVa 11374 Airport Express / Dijkstra

有2中种类型的路 第二种只能用一条 求出起点到所有点的最短路和终点到所有点的最短路 在枚举每一条路

输出3部分

打印路径

如果用了第二种类型的边 输出边的起点 没用输出Ticket Not Used

最短路

#include 
#include 
#include 
#include 
using namespace std;
const int maxn = 10510;
const int INF = 999999999;

struct HeapNode
{
	int d, u;
	bool operator < (const HeapNode& rhs) const
	{
		return d > rhs.d;
	}
};

struct Edge
{
	int from, to, dist;
};
vector  edges;
vector  edge2;
vector  G[maxn];
bool done[maxn];
int p[2][maxn];
int d[2][maxn];
int n, s, e, m, k;

void AddEdge(int from, int to, int dist)
{
	edges.push_back((Edge){from, to, dist});
	edges.push_back((Edge){to, from, dist});
	m = edges.size();
	G[from].push_back(m-2);
	G[to].push_back(m-1);
}
void AddEdge2(int from, int to, int dist)
{
	edge2.push_back((Edge){from, to, dist});
	edge2.push_back((Edge){to, from, dist});
}
void Dijkstra(int start, int num)
{
	priority_queue  Q;
	for(int i = 1; i <= n; i++)
		d[num][i] = INF;
	d[num][start] = 0;
	memset(done, 0, sizeof(done));
	Q.push((HeapNode){0, start});
	while(!Q.empty())
	{
		HeapNode x = Q.top();Q.pop();
		int u = x.u;
		if(done[u])
			continue;
		done[u] = true;
		//printf("%d\n", u);
		for(int i = 0; i < G[u].size(); i++)
		{
			Edge& e = edges[G[u][i]];
			if(d[num][e.to] > d[num][u] + e.dist)
			{
				d[num][e.to] = d[num][u] + e.dist;
				p[num][e.to] = G[u][i];
				Q.push((HeapNode){d[num][e.to], e.to});
			} 
		}
	}
}
int main()
{
	int cas = 0;
	while(scanf("%d %d %d", &n, &s, &e) != EOF)
	{
		if(cas++)
			puts("");
		edges.clear();
		edge2.clear();
		for(int i = 1; i <= n; i++)
			G[i].clear();
		scanf("%d", &k);
		for(int i = 1; i <= k; i++)
		{
			int u, v, w;
			scanf("%d %d %d", &u, &v, &w);
			AddEdge(u, v, w);
		}
		scanf("%d", &k);
		for(int i = 1; i <= k; i++)
		{
			int u, v, w;
			scanf("%d %d %d", &u, &v, &w);
			AddEdge2(u, v, w);
		}
		Dijkstra(s, 0);
		Dijkstra(e, 1);
		int ans = d[0][e];
		int flag = 0;
		int from;
		int to;
		for(int i = 0; i < edge2.size(); i++)
		{
			int temp = d[0][edge2[i].from] + d[1][edge2[i].to] + edge2[i].dist; 
			//printf("%d %d\n", d[1][3], d[0][1]);
			if(ans > temp)
			{
				flag = 1;
				from = edge2[i].from;
				to = edge2[i].to;
				ans = temp;
			}
		}
		int res1[10000];
		int res2[10000];
		int res3[10000];
		if(flag)
		{
			int pos = from, i = 0;
			while(pos != s)
			{	
				res1[i++] = edges[p[0][pos]].from;
				pos = edges[p[0][pos]].from;
			}
			int z = 0;
			for(int j = i-1; j >= 0; j--)
				res3[z++] = res1[j];
			res3[z++] = from;
			res3[z++] = to;
			pos = to, i = 0;
			while(pos != e)
			{	
				res2[i++] = edges[p[1][pos]].from;
				pos = edges[p[1][pos]].from;
			}
			
			for(int j = 0; j < i; j++)
				res3[z++] = res2[j];
			for(i = 0; i < z; i++)
			{
				if(i)
					printf(" ");
				printf("%d", res3[i]);
			}
			puts("");
		}
		else
		{
			int pos = e, i = 0;
			while(pos != s)
			{	
				res1[i++] = edges[p[0][pos]].from;
				pos = edges[p[0][pos]].from;
			}
			for(int j = i-1; j >= 0; j--)
				printf("%d ", res1[j]);
			printf("%d\n", e);
		}
		if(flag)
			printf("%d\n", from);
		else
			puts("Ticket Not Used");
		printf("%d\n", ans);
	}
	return 0;
}
/*
3 1 3
1
1 2 10
1
2 3 100
*/