# 【最短路·双Dijkstra】Silver Cow Party

 Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 30741 Accepted: 13936

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: NM, and X
Lines 2..M+1: Line i+1 describes road i with three space-separated integers: AiBi, 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.

1.邻接矩阵：(注意矩阵转置的做法)
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
#define maxn 1010
int map[maxn][maxn],n;
int way[maxn],dis[maxn];

void dijkstra(int x)
{
int visit[maxn],i,j,min,next=x;
memset(visit,0,sizeof(visit));
for(i=1;i<=n;i++)
dis[i]=map[x][i];
visit[x]=1;
for(i=2;i<=n;i++)
{
min=INF;
for(j=1;j<=n;++j)
{
if(!visit[j]&&dis[j]<min)
{
min=dis[j];
next=j;
}
}
visit[next]=1;
for(j=1;j<=n;++j)
{
if(!visit[j]&&dis[j]>dis[next]+map[next][j])
dis[j]=dis[next]+map[next][j];
}
}
}

int main()
{
int m,x,i,j,a,b,t;
while(cin>>n>>m>>x)
{
for(i=1;i<=n;i++)
{
for(j=1;j<=n;++j)
{
if(i!=j)
map[i][j]=INF;
else
map[i][j]=0;
}
}
while(m--)
{
scanf("%d%d%d",&a,&b,&t);
map[a][b]=t;
}
dijkstra(x);
for(i=1;i<=n;i++)//存下来之前的结果
way[i]=dis[i];
int ans=0;
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;++j)
swap(map[i][j],map[j][i]);
}
dijkstra(x);
for(i=1;i<=n;i++)
{
if(i!=x)
ans=max(ans,way[i]+dis[i]);
}
printf("%d\n",ans);
}
return 0;
}

2.堆优化的Dijkstra：(注意别忘了全面初始化)

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=100000;
struct qnode
{
int v;
int c;
qnode(int v=0,int c=0):v(v),c(c){}
bool operator <(const qnode &r)const
{
return c>r.c;
}
};
struct store
{
int u,v,w;
};
vector<store> re;//反向存图

struct Edge
{
int v,cost;
Edge(int v=0,int cost=0):v(v),cost(cost){}
};
vector<Edge>E[MAXN];//图

bool vis[MAXN];
int dist[MAXN];
void Dijkstra(int n,int start)//点的编号从1开始
{
memset(vis,false,sizeof(vis));
for(int i=1;i<=n;i++)dist[i]=INF;
priority_queue<qnode>que;
while(!que.empty())que.pop();//初始化
dist[start]=0;
que.push(qnode(start,0));
qnode tmp;
while(!que.empty())
{
tmp=que.top();
que.pop();
int u=tmp.v;
if(vis[u])continue;
vis[u]=true;
for(int i=0;i<E[u].size();i++)
{
int v=E[tmp.v][i].v;
int cost=E[u][i].cost;
if(!vis[v]&&dist[v]>dist[u]+cost)
{
dist[v]=dist[u]+cost;
que.push(qnode(v,dist[v]));
}
}
}
}
{
E[u].push_back(Edge(v,w));
}
int main()
{
int n,m,x;
int way_cost[100000];
while(cin>>n>>m>>x)
{
for(int i=1;i<=n;i++) E[i].clear();
for(int i=0;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
store temp;
temp.u=v;
temp.v=u;
temp.w=w;
re.push_back(temp);
}
Dijkstra(n,x);
for(int i=1;i<=m;i++)
way_cost[i]=dist[i];
E->clear();
for(int i=1;i<=n;i++) E[i].clear();

for(int i=0;i<m;i++)
{
}
Dijkstra(n,x);
int ans=-1;
for(int i=1;i<=n;i++)
{
if(i!=x)
if(ans<way_cost[i]+dist[i])
ans=way_cost[i]+dist[i];
}
printf("%d\n",ans);
}
return 0;
}