次短路

题目链接

poj3255

Roadblocks

Time Limit: 2000MS
Memory Limit: 65536K

Description

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).

Input

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)
Output

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

Sample Input

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

Sample Output

450

Hint

Two routes: 1 -> 2 -> 4 (length 100+200=300) and 1 -> 2 -> 3 -> 4 (length 100+250+100=450)

解题思路

跟求次小值类似，即：更新最短路的同时更新次短路，再更新两次次短路：次短路与前一次次短路，当前没有被更新的最短路与次短路~
以spfa算法为例，其他算法最短路模板同理改动即可~

• 时间复杂度：$O(kn)$

代码

#include<queue>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=5010;
int n,m;
vector<pair<int,int> > adj[N];
int d1[N],d2[N];
bool v[N];
void spfa()
{
    memset(v,0,sizeof v);
    memset(d1,0x3f,sizeof d1);
    memset(d2,0x3f,sizeof d2);
    queue<int> q;
    q.push(1);
    d1[1]=0;
    v[1]=true;
    while(q.size())
    {
        int x=q.front();
        q.pop();
        v[x]=false;
        for(int i=0;i<adj[x].size();i++)
        {
            int y=adj[x][i].first,w=adj[x][i].second;
            if(d1[y]>d1[x]+w)
            {
                d2[y]=d1[y];
                d1[y]=d1[x]+w;
                if(!v[y])
                    q.push(y),v[y]=true;
            }
            if(d2[y]>d2[x]+w)
            {
                d2[y]=d2[x]+w;
                if(!v[y])
                    q.push(y),v[y]=true;
            }
            if(d1[y]<d1[x]+w&&d2[y]>d1[x]+w)
            {
                d2[y]=d1[x]+w;
                if(!v[y])
                    q.push(y),v[y]=true;
            }
        }
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    while(m--)
    {
        int x,y,w;
        scanf("%d%d%d",&x,&y,&w);
        adj[x].push_back(make_pair(y,w));
        adj[y].push_back(make_pair(x,w));
    }
    spfa();
    printf("%d",d2[n]);
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/15365328.html
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