学步园

大意：不再赘述，牛牛们一定可以到达目的地，当然也可以走回去。

思路：建立正、反向图，使用spfa,Dijkstra，枚举最大值均可。

CODE：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib> 
#include <queue>
using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 1010;
const int MAXM = 100010;

struct Edge
{
	int v, next, w;
}edge[MAXM];

int n, m, src;
int cnt;

int first[MAXN], d[MAXN], ans[MAXN];
int su[MAXM], sv[MAXM], sw[MAXM];
int MaxTime;

inline void read_graph(int u, int v, int w)
{
	edge[cnt].v = v;
	edge[cnt].w = w;
	edge[cnt].next = first[u], first[u] = cnt++;
}

inline void init()
{
	cnt = 0;
	MaxTime = 0;
	memset(first, -1, sizeof(first));
}

void spfa(int src)
{
	queue<int> q;
	bool inq[MAXN] = {0};
	for(int i = 1; i <= n; i++) d[i] = (i == src)? 0:INF;
	q.push(src);
	while(!q.empty())
	{
		int x = q.front(); q.pop();
		inq[x] = 0;
		for(int e = first[x]; e != -1; e = edge[e].next)
		{
			int v = edge[e].v, w = edge[e].w;
			if(d[v] > d[x] + w)
			{
				d[v] = d[x] + w;
				if(!inq[v])
				{
					inq[v] = 1;
					q.push(v);
				}
			}
		}
	}
}

void solve()
{
	init();
	for(int i = 0; i < m; i++)
	{
		scanf("%d%d%d", &su[i], &sv[i], &sw[i]);
		read_graph(su[i], sv[i], sw[i]);
	}
	spfa(src);
	for(int i = 1; i <= n; i++) ans[i] += d[i];
	init();
	for(int i = 0; i < m; i++) read_graph(sv[i], su[i], sw[i]);
	spfa(src);
	for(int i = 1; i <= n; i++)
	{
		ans[i] += d[i];
		MaxTime = max(MaxTime, ans[i]);
	}
	printf("%d\n", MaxTime);
}

int main()
{
	while(~scanf("%d%d%d", &n, &m, &src))
	{
		solve();
	}
	return 0;
}