PAT A 1030. Travel Plan (30)【最短路徑】

https://www.patest.cn/contests/pat-a-practise/1030

找最短路，如果有多條找最小消耗的，相當於找兩次最短路，可以直接dfs，資料小不會超時。

#include<cstdio>  
#include<string>  
#include<cstring>  
#include<vector>  
#include<iostream>  
#include<queue>  
#include<bitset>  
#include<algorithm>  
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 5e2 + 10;
int n, m, s, t, map[maxn][maxn], cost[maxn][maxn], x, y, z, c;
int dis[maxn], v[maxn];

void dfs(int x)
{
	if (x == t)return;
	for (int i = 0; i < n; i++)
	{
		if (map[x][i])
		{
			if (dis[i]>dis[x]+map[x][i])
			{
				dis[i] = dis[x] + map[x][i];
				v[i] = v[x] + cost[x][i];
				dfs(i);
			}
			else if (dis[i] == dis[x] + map[x][i] && v[i] > v[x] + cost[x][i])
			{
				v[i] = v[x] + cost[x][i];
				dfs(i);
			}
		}
	}
}

bool Dfs(int x)
{
	if (x == s){ printf("%d ", s); return true; }
	for (int i = 0; i < n; i++)
	{
		if (map[x][i] && dis[x] == dis[i] + map[x][i] && v[x] == v[i] + cost[x][i])
		{
			if (Dfs(i)){printf("%d ", x); return true;}
		}
	}
	return false;
}

int main()
{
	scanf("%d%d%d%d", &n, &m, &s, &t);
	while (m--)
	{
		scanf("%d%d%d%d", &x, &y, &z, &c);
		if (!map[x][y] || map[x][y] > z)
		{
			map[x][y] = map[y][x] = z;
			cost[x][y] = cost[y][x] = c;
		}
		else if (map[x][y] == z) cost[x][y] = cost[y][x] = min(z, cost[x][y]);
	}
	for (int i = 0; i < n; i++)dis[i] = v[i] = INF;
	dis[s] = v[s] = 0;
	dfs(s);
	Dfs(t);
	printf("%d %d\n", dis[t], v[t]);
	return 0;
}