oooccc1 is a Software Engineer who has to ride to the work place every Monday through Friday. For a long period, he went to office with the shortest path because he loves to sleep late…Time goes by, he find that he should have some changes as you could see, always riding with the same path is boring.
One day, oooccc1 got an idea! Why could I take another path? Tired at all the tasks he got, he got no time to carry it out. As a best friend of his, you’re going to help him!
Since oooccc1 is now getting up earlier, he is glad to take those paths, which are a little longer than the shortest one. To be precisely, you are going to find all the second shortest paths.
You would be given a directed graph G, together with the start point S which stands for oooccc’1 his house and target point E presents his office. And there is no cycle in the graph. Your task is to tell him how long are these paths and how many there are.
Input
There are some cases. Proceed till the end of file.
The first line of each case is three integers N, M, S, E (3 <= N <= 50, 0 <= S , E <N)
N stands for the nodes in that graph, M stands for the number of edges, S stands for the start point, and E stands for the end point.
Then M lines follows to describe the edges: x y w. x stands for the start point, and y stands for another point, w stands for the length between x and y.
All the nodes are marked from 0 to N-1.
Output
For each case,please output the length and count for those second shortest paths in one line. Separate them with a single space.
Sample Input
3 3 0 2
0 2 5
0 1 4
1 2 2Sample Output
6 1题解：题目已经说明题意了；我们可以记录最短路和短路的长度及其次数并不断更新他们。每次更新是无非5种情况：比最小值小，等于最小值，大于最小值小于次小值，等于次小值，大于次小值；
AC代码为：
/*
题意：给你N个点M条有向边，开始点s，终点e，求 s到e的次最短路，输出次最短路的长度和条数
*/
#include<bits/stdc++.h>
using namespace std;
#define MAX 1100
#define INF 999999999
struct edge
{
int from,to,w;
};
vector<edge> edges;
vector<int> G[MAX];
int m,n;
int dis[MAX][2],vis[MAX][2],cnt[MAX][2];
int s,e;
void addedge(int x,int y,int w)
{
edge a={x,y,w};
edges.push_back(a);
G[x].push_back(edges.size()-1);
}
void dijkstra(int x,int y)
{
for(int i=0;i<=n;i++)
{
dis[i][0]=dis[i][1]=INF;
cnt[i][0]=cnt[i][1]=INF;
}
dis[x][0]=0;
cnt[x][0]=1;
memset(vis,0,sizeof(vis));
for(int i=1;i<=n*2;i++)
{
int min=INF,u=-1,flag;
for(int j=0;j<n;j++)
{
if(!vis[j][0]&&min>dis[j][0])
{
min=dis[j][0];
flag=0;
u=j;
}
else if(!vis[j][1]&&min>dis[j][1])
{
min=dis[j][1];
flag=1;
u=j;
}
}
if(u==-1) break;
vis[u][flag]=1;
for(int j=0;j<G[u].size();j++)
{
edge v=edges[G[u][j]];
if(min+v.w<dis[v.to][0])
{
dis[v.to][1]=dis[v.to][0];
cnt[v.to][1]=cnt[v.to][0];
dis[v.to][0]=min+v.w;
cnt[v.to][0]=cnt[u][flag];
}
else if(min+v.w==dis[v.to][0]) cnt[v.to][0]+=cnt[u][flag];
else if(min+v.w==dis[v.to][1]) cnt[v.to][1]+=cnt[u][flag];
else if(min+v.w<dis[v.to][1])
{
dis[v.to][1]=min+v.w;
cnt[v.to][1]=cnt[u][flag];
}
}
}
printf("%d %d
",dis[e][1],cnt[e][1]);
}
int main()
{
while(scanf("%d %d %d %d",&n,&m,&s,&e)!=EOF)
{
for(int i=0;i<=n;i++) G[i].clear();
edges.clear();
for(int i=1;i<=m;i++)
{
int x,y,w;
scanf("%d %d %d",&x,&y,&w);
addedge(x,y,w);
}
dijkstra(s,e);
}
return 0;
}