奶牛开会 最短路

Description

One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤
X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road
i requires Ti (1 ≤ Ti ≤ 100) units of time to traverse.

Each cow must walk to the party and, when the party is over, return to her farm. Each cow is lazy and thus picks an optimal route with the shortest time. A cow's return route might be different from her original route to the
party since roads are one-way.

Of all the cows, what is the longest amount of time a cow must spend walking to the party and back?

Input

Line 1: Three space-separated integers, respectively:
N, M, and X

Lines 2.. M+1: Line i+1 describes road i with three space-separated integers:
Ai, Bi, and Ti. The described road runs from farm
Ai to farm Bi, requiring Ti time units to traverse.

Output

Line 1: One integer: the maximum of time any one cow must walk.

Sample Input

4 8 2
1 2 4
1 3 2
1 4 7
2 1 1
2 3 5
3 1 2
3 4 4
4 2 3

Sample Output

10

Hint

Cow 4 proceeds directly to the party (3 units) and returns via farms 1 and 3 (7 units), for a total of 10 time units.

英语不好╮(╯﹏╰）╭翻译不靠谱 自己yy一下

题意：

解释你说下样例

第一行4个点 8条边 到2开会（终点）

 算1到2 加2 到 1的权值和，与3 到 2加 2到3的权值和，4到2加2到4的权值和 ，取最大值输出。

PS：数组别开太大，重点反向建边。

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
using namespace std;
int V,E,s,t,tot;
const int size = 20000;
int first[size],next[size];
long long dis[size];
bool used[size];
long long ans[10000];
long long tt[1000];
struct edge
{
int from;
int to;
long long d;
<
dcbd
span class="sh-cbracket">}es[size];
struct date
{
int f,t,d;
}in[size];
void jt(int f,int t,int d)
{
es[++tot]=(edge){f,t,d};
next[tot]=first[f];
first[f]=tot;
}
queue <int> q;
void spfa(int s)
{
for(int i=1;i<=V;i++)
dis[i]=2147483641234567ll;
dis[s]=0;
q.push(s);
used[s]=1;
while (!q.empty())
{
int u=q.front();
q.pop();
used[u]=0;
for(int i=first[u];i!=0;i=next[i])
{
int v=es[i].to;
if(dis[v]>dis[u]+es[i].d)
{
dis[v]=dis[u]+es[i].d;
if(used[v]==0)
{
q.push(v);
used[v]=1;
}
}
}
}
}
void init()
{
memset(es,0,sizeof(es));
memset(first,0,sizeof(first));
memset(next,0,sizeof(next));
tot = 0;
}
int read()
{
int x = 0 , f = 1;
char in = getchar();
while(in < '0' || in > '9')
{
if(in == '-')
f = -1;
in = getchar();
}
while(in >= '0' && in <= '9')
{
x = x * 10 + in - '0';
in = getchar();
}
return x * f;
}
int main()
{

memset(in,0,sizeof(in));
init();
int ss;
V = read() , E = read(),ss=read();
for(int i=1;i<=E;i++)
{
in[i].f = read() , in[i].t =  read() , in[i].d = read();
jt(in[i].f,in[i].t,in[i].d);
}
spfa(ss);

for(int i=1;i<=V;i++)
{
ans[i]+=dis[i];
}
init();
for(int i=1;i<=E;i++)
jt(in[i].t,in[i].f,in[i].d);
spfa(ss);
for(int i=1;i<=V;i++)
{
ans[i] += dis[i];
}
int da=-1;
for(int i=1;i<=V;i++)
if(ans[i]>da)
da=ans[i];
printf("%d\n",da);
return 0;
}