路障【SPFA】

题目大意：

求一个无向图的次短路。
Input$Input$

4 4
1 2 100
2 4 200
2 3 250
3 4 100

Output$Output$

450

---

思路：

这道题，正解是跑两遍SPFA$SPFA$，一遍是从点1$1$，求出到达其他点的最短路径，记作dis1[i]$dis1[i]$；再从点n$n$，也求出到达其他点的最短路径，记作dis2[i]$dis2[i]$；最后枚举每条边，若两个端点分别是u[i]$u[i]$和v[i]$v[i]$，长度为dis[i]$dis[i]$，那么就是求次小的

dis[i]+dis1[u[i]]+dis2[v[i]]$$dis[i]+dis1[u[i]]+dis2[v[i]]$$

或

dis[i]+dis2[u[i]]+dis1[v[i]]$$dis[i]+dis2[u[i]]+dis1[v[i]]$$

---

代码：

#include <cstdio>
#include <iostream>
#include <queue>
#define fre(x) freopen(#x".in","r",stdin),freopen(#x".out","w",stdout);
using namespace std;

const int Inf=99999999;
int n,m,x,y,z,t,dis1[30001],dis2[30001],vis[30001],head[30001],dis[300001],X[300001],Y[300001],ans,minn,sum;

struct edge  //邻接表
{
    int next,to,dis;
}e[300001];

void add(int from,int to,int d)  //连边
{
    t++;
    e[t].to=to;
    e[t].dis=d;
    e[t].next=head[from];
    head[from]=t;
}

void spfa1()  //从点1开始跑SPFA
{
    queue<int> q;
    for (int i=1;i<=n;i++)
    {
        vis[i]=0;
        dis1[i]=Inf;
    }
    vis[1]=1;
    dis1[1]=0;
    q.push(1);
    while (q.size())  //队列不为空
    {
        int u=q.front();
        vis[u]=0;
        q.pop();
        for (int i=head[u];i;i=e[i].next)
        {
            int v=e[i].to;
            if (dis1[v]>dis1[u]+e[i].dis)  //更新最短路
            {
                dis1[v]=dis1[u]+e[i].dis;
                if (!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
}

void spfa2()  //从点n开始跑SPFA
{
    queue<int> q;
    for (int i=1;i<=n;i++)
    {
        vis[i]=0;
        dis2[i]=Inf;
    }
    vis[n]=1;
    dis2[n]=0;
    q.push(n);
    while (q.size())
    {
        int u=q.front();
        vis[u]=0;
        q.pop();
        for (int i=head[u];i;i=e[i].next)
        {
            int v=e[i].to;
            if (dis2[v]>dis2[u]+e[i].dis)
            {
                dis2[v]=dis2[u]+e[i].dis;
                if (!vis[v])
                {
                    q.push(v);
                    vis[v]=1;
                }
            }
        }
    }
}

int main()
{
    fre(block);
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        dis[i]=z;
        X[i]=x;
        Y[i]=y;  //记录
        add(x,y,z);
        add(y,x,z);  //连边
    }
    spfa1();
    spfa2();
    minn=ans=Inf;
    for (int i=1;i<=m;i++)  //求次短路
    {
        sum=dis[i]+dis1[X[i]]+dis2[Y[i]];
        if (sum<minn)
        {
            ans=minn;
            minn=sum;
        }
        else if (sum<ans&&sum!=minn) ans=sum;
        sum=dis[i]+dis2[X[i]]+dis1[Y[i]];
        if (sum<minn)
        {
            ans=minn;
            minn=sum;
        }
        else if (sum<ans&&sum!=minn) ans=sum;
    }
    printf("%d\n",ans);
    return 0;
}