How Many Paths Are There

3 3 0 2
0 2 5
0 1 4
1 2 2

6 1

参考代码：

#include<stack>
#include<queue>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const double eps=1e-6;
const int INF=0x3f3f3f3f;
const int MAXN=52;

struct point
{
    int pos,cost, ismin;
    bool operator < (const point &p) const//每次从优先队列中取出集合中最短路最大的点
    {
        if(p.cost!=cost)
            return p.cost<cost;
        return p.pos<pos;
    }
    point(int pos,int cost,int ismin)
    {
        this->pos=pos;
        this->cost=cost;
        this->ismin=ismin;
    }
};

int n,m,s,e,mincost[MAXN][3],dp[MAXN][3],edge[MAXN][MAXN];
bool used[MAXN][3];

void SPFA()
{
    memset(mincost,INF,sizeof(mincost));
    memset(dp,0,sizeof(dp));
    memset(used,false,sizeof(used));
    priority_queue<point>q;//优先队列
    mincost[s][1]=0;
    dp[s][1]=1;
    q.push(point(s,0,1));
    while(!q.empty())
    {
        point now=q.top();
        q.pop();
        if(used[now.pos][now.ismin])continue;
        used[now.pos][now.ismin]=true;
        for(int i=0; i<n; i++)
        {
            if(edge[now.pos][i]==INF)
                continue;
            if(!used[i][1]&&now.cost+edge[now.pos][i]<mincost[i][1])//从当前点出发到达i的某条路的长度比到达i点的最短路短
            {
                mincost[i][2]=mincost[i][1];//更新次短路
                dp[i][2]=dp[i][1];
                q.push(point(i,mincost[i][2],2));
            
                mincost[i][1]=now.cost+edge[now.pos][i];//更新最短路
                dp[i][1]=dp[now.pos][now.ismin];
                q.push(point(i,mincost[i][1],1));
            }
            else if(!used[i][1]&&now.cost+edge[now.pos][i]==mincost[i][1])//从当前点出发到达i的某条路的长度与到达i点的最短路的长度一致
            {
                dp[i][1]+=dp[now.pos][now.ismin];
            }
            else if(!used[i][2]&&now.cost+edge[now.pos][i]<mincost[i][2])//从当前点出发到达i的某条路的长度与到达i点的最短路长，但同时比次短路短
            {
                mincost[i][2]=now.cost+edge[now.pos][i];
                dp[i][2]=dp[now.pos][now.ismin];
                q.push(point(i,mincost[i][2],2));
            }
            else if(!used[i][2]&&now.cost+edge[now.pos][i]==mincost[i][2])//从当前点出发到达i的某条路的长度与到达i点的最次路的长度一致
            {
                dp[i][2]+=dp[now.pos][now.ismin];
            }
        }
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
    while(scanf("%d%d%d%d",&n,&m,&s,&e)!=EOF)
    {
        memset(edge,INF,sizeof(edge));
        for(int i=1; i<=m; i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            edge[a][b]=c;//有向图
        }
        SPFA();
        printf("%d %d\n",mincost[e][2],dp[e][2]);
    }
    return 0;
}

HDU 3191 How Many Paths Are There（SPFA）

原文：http://blog.csdn.net/noooooorth/article/details/47319347