洛谷题解 P2865 【[USACO06NOV]路障Roadblocks】

链接：https://www.luogu.org/problemnew/show/P2865

题目描述

Bessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has decided to take the second-shortest rather than the shortest path. She knows there must be some second-shortest path.

The countryside consists of R (1 ≤ R ≤ 100,000) bidirectional roads, each linking two of the N (1 ≤ N ≤ 5000) intersections, conveniently numbered 1..N. Bessie starts at intersection 1, and her friend (the destination) is at intersection N.

The second-shortest path may share roads with any of the shortest paths, and it may backtrack i.e., use the same road or intersection more than once. The second-shortest path is the shortest path whose length is longer than the shortest path(s) (i.e., if two or more shortest paths exist, the second-shortest path is the one whose length is longer than those but no longer than any other path).

贝茜把家搬到了一个小农场，但她常常回到FJ的农场去拜访她的朋友。贝茜很喜欢路边的风景，不想那么快地结束她的旅途，于是她每次回农场，都会选择第二短的路径，而不象我们所习惯的那样，选择最短路。 贝茜所在的乡村有R(1<=R<=100,000)条双向道路，每条路都联结了所有的N(1<=N<=5000)个农场中的某两个。贝茜居住在农场1，她的朋友们居住在农场N（即贝茜每次旅行的目的地）。 贝茜选择的第二短的路径中，可以包含任何一条在最短路中出现的道路，并且，一条路可以重复走多次。当然咯，第二短路的长度必须严格大于最短路（可能有多条）的长度，但它的长度必须不大于所有除最短路外的路径的长度。

输入输出格式

输入格式：

Line 1: Two space-separated integers: N and R

Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

输出格式：

Line 1: The length of the second shortest path between node 1 and node N

 

题解：

在做本题前，一定理解最短路算法的核心思想。 当然，并不需要对所有的最短路算法有深刻了解。 所以今天我就给大家带来优先队列优化的dijsktra。（SPFA我不会）

那么问题来了，经典的dijkstra只能维护最短路，怎么去维护次短路呢？其实很简单，只要同时维护两个数组就行了。

那怎么更新呢？我们设dis1数组存最短路，dis2数组存最长路，当找到一条路，比dis1和dis2都短时，就先把目前次短路赋为目前最短路，也就是让dis2=dis1，再把目前最短路赋为找到的更短路。剩下的就是dijkstra的板子了。

当我们存边时，还是推荐大家用邻接表存边，只用写一个函数，跑的好像还比vector快，下面就为大家带来一个不怎么好看的dijkstra板：

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<stack>
#include<vector>
#include<cstdio>
#define ll long long
#define inf 1e9+10
using namespace std;
const int maxn=1e5+10;
struct mxr
{
    int to,next,val;
}e[2*maxn];
int n,cnt,m,vis[maxn],head[maxn],s;
int dis[maxn];
priority_queue<pair<ll,int> >q;
inline int read()
{
    int s=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){ch=getchar();
    if(ch=='-')f=-1;}
    while(ch>='0'&&ch<='9')
    s=(s<<3)+(s<<1)+(ch^48),ch=getchar();
    return s*f;
}
inline void add(int u,int v,int w)
{
    e[cnt].to=v;
    e[cnt].val=w;
    e[cnt].next=head[u];
    head[u]=cnt++;
    return ;
}
inline void dij()
{
    q.push(make_pair(0,s));
    dis[s]=0;
    while(!q.empty())
    {
        int now=q.top().second;
        q.pop();
        if(vis[now]) continue;
        vis[now]=1;
        for(int i=head[now];i!=-1;i=e[i].next)
        {
            int v=e[i].to;
            if(dis[v]>dis[now]+e[i].val)
            {
                dis[v]=dis[now]+e[i].val;
                q.push(make_pair(-dis[v],v));
            }
        }
    }
    return ;
}
int main()
{
    memset(head,-1,sizeof(head));
    cnt=0;
    n=read(),m=read(),s=read();
    for(int i=1;i<=n;i++) dis[i]=inf;
    for(int i=1;i<=m;i++)
    {
        int a,b,w;
        a=read(),b=read(),w=read();
        add(a,b,w);
    }
    dij();
    for(int i=1;i<=n;i++) printf("%d ",dis[i]);
    return 0;
}

下面是本题代码：

#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<stack>
#include<vector>
#include<cstdio>
#define ll long long
#define inf 1e9+10
using namespace std;
const int maxn=1e5+10;
struct mxr
{
    int to,next,val;
}e[2*maxn];
int n,cnt,m,head[maxn];
int dis1[maxn],dis2[maxn];
priority_queue<pair<ll,int> >q;
inline char nc()//fread，跑的快 
{
    static char buf[100000],*L=buf,*R=buf;
    return L==R&&(R=(L=buf)+fread(buf,1,100000,stdin),L==R)?EOF:*L++;
}
inline int read()
{
    int ret=0,f=1;char ch=nc();
    while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=nc();}
    while (ch>='0'&&ch<='9') ret=ret*10+ch-'0',ch=nc();
    return ret*f;
}
inline void write(int x)
{
    if(x<0) putchar('-'),x=-x;
    if(x>9) write(x/10);
    putchar(x%10+'0');
    return ;
}
inline void add(int u,int v,int w)//邻接表存边，貌似比vector快 
{
    e[cnt].to=v;
    e[cnt].val=w;
    e[cnt].next=head[u];
    head[u]=cnt++;
    return ;
}
inline void dij()//优先队列优化dijkstra 
{
    for(register int i=1;i<=n;++i) dis1[i]=inf,dis2[i]=inf;
    dis1[1]=0;
    q.push(make_pair(0,1));
    while(!q.empty())
    {
        int now=q.top().second,dis=-q.top().first;
        q.pop();
        for(register int i=head[now];i!=-1;i=e[i].next)
        {
            int v=e[i].to;
            if(dis1[v]>dis+e[i].val)
                dis2[v]=dis1[v],dis1[v]=dis+e[i].val,q.push(make_pair(-dis1[v],v));//找到更优，更新dis1与dis2 
            if(dis2[v]>dis+e[i].val&&dis1[v]<dis+e[i].val)
                dis2[v]=dis+e[i].val,q.push(make_pair(-dis2[v],v));//找到比dis2优，比dis1劣，更新dis2 
        }
    }
    return ;
}
int main()
{
    memset(head,-1,sizeof(head));
    cnt=0;
    n=read(),m=read();
    for(register int i=1;i<=m;++i)
    {
        int a,b,w;
        a=read(),b=read(),w=read();
        add(a,b,w),add(b,a,w);
    }
    dij();
    write(dis2[n]),puts(" ");
    return 0;
}

谢谢大家！