【PAT-A1030】Travel Plan（最短路徑）

【分析】

典型的最短路徑問題。以邊權作爲第二標尺，要求所有最短路徑中第二標尺和最小的路徑。
注意點就是Dijkstra和DFS函數的書寫。
Dijkstra裏執行n次的總循環不要弄錯了。

【代碼】

#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 510;
const int INF = (1 << 30) - 1;
int G[maxn][maxn], d[maxn], cost[maxn][maxn];
int c1, c2, dis, cst;
int n, m, st, en;
bool vis[maxn];
vector<int> pre[maxn], path, tmpPath;
int minCost = INF;
void Dijkstra(int st) {
	fill(d, d + maxn, INF);
	fill(vis, vis + maxn, false);
	d[st] = 0;
	
	for (int i = 0; i < n; i++)
	{
		int u = -1, MIN = INF;
		for (int j = 0; j < n; j++)
		{
			if (vis[j] == false && d[j] < MIN) {
				MIN = d[j];
				u = j;
			}
		}

		if (u == -1) return;
		vis[u] = true;
		for (int v = 0; v < n; v++)
		{
			if (vis[v] == false && G[u][v] != INF) {// 記得給G[][]初始化
				if (d[v] > d[u] + G[u][v]) {
					d[v] = d[u] + G[u][v];
					pre[v].clear();
					pre[v].push_back(u);
				}
				else if (d[v] == d[u] + G[u][v]) {
					pre[v].push_back(u);
				}
			}
		}
	}
}

void DFS(int v, int st) {
	if (v == st) {
		tmpPath.push_back(v);
		//計算最小權值和
		int costSum = 0;
		for (int i = 0; i < tmpPath.size() - 1; i++)
		{
			costSum += cost[tmpPath[i]][tmpPath[i + 1]];
		}
		if (costSum < minCost) {
			minCost = costSum;
			path = tmpPath;
		}
		tmpPath.pop_back();
		return;
	}

	tmpPath.push_back(v);
	for (int i = 0; i < pre[v].size(); i++)
	{
		DFS(pre[v][i], st);
	}
	tmpPath.pop_back();
}
int main() {
	fill(G[0], G[0] + maxn * maxn, INF);
	fill(cost[0], cost[0] + maxn * maxn, INF);
	scanf("%d%d%d%d", &n, &m, &st, &en);
	for (int i = 0; i < m; i++)
	{
		scanf("%d%d%d%d", &c1, &c2, &dis, &cst);
		G[c1][c2] = G[c2][c1] = dis;
		cost[c1][c2] = cost[c2][c1] = cst;
	}
	Dijkstra(st);
	DFS(en, st);
	for (int i = path.size() - 1; i >=0; i--)
	{
		printf("%d ", path[i]);
	}
	printf("%d %d\n", d[en], minCost);
	return 0;
}