题意：给定n个点（点标[0,n-1]）m条有向边，起点和终点
问次短路的条数（一定存在次短路权值>最短路)
记录每个点的最短路，次短路的长度和方法数
则当更新最短路时就要把旧的最短路路径赋给次短路，剩下的状态转移还是比较明显的
而所有可转移的点中优先转移先更新最近的点。则如此更新有些类似bfs更新过就不再更新该点。
因为给出的图中所有边{u,v,dis}，u<v ;
必然无环且有一定偏序关系，则必须先更新点标小的点。
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
#include <set>
#include <vector>
using namespace std;
#define ll int
#define inf 100000000
#define N 55
struct Edge{
int from, to, dis, nex;
}edge[N*N*2];
int head[N], edgenum ;
void add(int u, int v, int d){
Edge E={u,v,d,head[u]};
edge[edgenum] = E;
head[u] = edgenum++;
}
int dis[N][2], dp[N][2];
int n, m, s, t;
bool inq[N][2];
struct node{
int to, flag, dis;
node(int a=0,int b=0,int c=0):to(a),flag(b),dis(c){}
bool operator<(const node&a)const{
if(a.dis==dis)return a.to<to;
return a.dis<dis;
}
};
void spfa(){
int i, j;
memset(dp, 0, sizeof(dp));
memset(inq, 0, sizeof(inq));
for(j=0;j<=n;j++)dis[j][0] = dis[j][1] = inf;
priority_queue<node>q; q.push(node(s,0,0));
dis[s][0] = 0; dp[s][0] = 1;
while(!q.empty()){
node u = q.top(); q.pop();
if(inq[u.to][u.flag])continue;
inq[u.to][u.flag] = 1;
for(int i = head[u.to]; ~i; i = edge[i].nex){
int v = edge[i].to;
int nowd = u.dis+edge[i].dis;
if(!inq[v][0] && nowd < dis[v][0])
{
if(dis[v][0]<inf)
{
dis[v][1] = dis[v][0];
dp[v][1] = dp[v][0];
q.push(node(v,1,dis[v][1]));
}
dis[v][0] = nowd;
dp[v][0] = dp[u.to][u.flag];
q.push(node(v,0,dis[v][0]));
}
else if(!inq[v][0] && nowd == dis[v][0])
{
dp[v][0] += dp[u.to][u.flag];
}
else if(!inq[v][1] && nowd < dis[v][1])
{
dis[v][1] = nowd;
dp[v][1] = dp[u.to][u.flag];
q.push(node(v,1,dis[v][1]));
}
else if(!inq[v][1] && nowd == dis[v][1])
{
dp[v][1] += dp[u.to][u.flag];
}
}
}
}
void init(){memset(head, -1, sizeof(head));edgenum = 0;}
int main(){
ll u, v, d;
while(~scanf("%d %d %d %d",&n,&m,&s,&t)){
init();
while(m--){
scanf("%d %d %d",&u,&v,&d);
add(u,v,d);
}
spfa();
printf("%d %d\n", dis[t][1], dp[t][1]);
}
return 0;
}
/*
3 3 0 2
0 2 5
0 1 4
1 2 2
8 28 0 7
0 1 1
0 2 2
0 3 3
0 4 4
0 5 5
0 6 6
0 7 6
1 2 1
1 3 2
1 4 3
1 5 4
1 6 5
1 7 6
2 3 1
2 4 2
2 5 3
2 6 4
2 7 5
3 4 1
3 5 2
3 6 3
3 7 4
4 5 1
4 6 2
4 7 3
5 6 1
5 7 2
6 7 1
5 6 0 4
0 1 1
0 2 1
2 1 0
1 3 0
1 4 1
3 4 0
6 8 0 4
0 1 1
0 2 1
2 1 0
1 3 0
1 4 1
3 4 0
2 5 0
5 1 0
*/
