【Dijkstra算法】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)

 

OJ:POJ

 

题目大意

某街区共有R条道路、N个路口。道路可以双向通行。问1号路口到N号路口的次短路长度是多少？次短路指的是比最短路长度长的次短的路径。同一条边可以经过多次。（摘自《挑战程序设计 第二版》）

 

分析

用Dijkstra算法来求，单源最短路问题，到某个顶点v的次短路要么是①到其他某个顶点u的最短路再加上u->v的边，要么是到u的次短路再加上u->v的边。

 

参考代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <queue>

using namespace std;

const int MAX_V=5001;
const int INF=1e9;

struct edge{
int to;//边的出点
int cost;//权值
};

vector<edge> G[MAX_V];//邻接表

typedef pair<int,int> P;//first是源点到该点的距离，second是当前顶点

int dist1[MAX_V];//存最短距离
int dist2[MAX_V];//存次短距离
int N,R;//N个路口,R条道路
void dijkstra();
int main()
{

    int s,w,t;
    edge e;
    scanf("%d%d",&N,&R);
    for(int i=0;i<R;i++){
    scanf("%d%d%d",&s,&t,&w);
    e.to=t-1;
    e.cost=w;
    G[s-1].push_back(e);
    e.to=s-1;
    G[t-1].push_back(e);
    }
    dijkstra();
    printf("%d
",dist2[N-1]);
    return 0;
}

void dijkstra(){
    priority_queue <P,vector<P>,greater<P> >que;
    fill(dist1,dist1+N,INF);
    fill(dist2,dist2+N,INF);
    dist1[0]=0;
    que.push(P(0,0));
    while(!que.empty()){
        P p=que.top();
        que.pop();
        int v=p.second;//新的源点
        int d=p.first;
        /*如果旧源点到新源点的距离比旧源点到新源点的距离大，
        那么不用执行下面的代码去更新新源点到其他各点的dist1,dist2
        因为它算出来的距离肯定比以前算出来的大。
        */
        if(d>dist2[v])
            continue;
        for(int i=0;i<G[v].size();i++){
            edge& e=G[v][i];
            int d2=d+e.cost;
            if(dist1[e.to]>d2){///最短距离
                swap(d2,dist1[e.to]);
                que.push(P(dist1[e.to], e.to));
            }
            /*
            d2大于源点到e.to的最短距离，小于以前计算的次短距离则更新
            */
            if(d2>dist1[e.to]&&d2<dist2[e.to]){///次短距离
                dist2[e.to]=d2;
                que.push(P(dist2[e.to],e.to));
            }
        }

    }

}

参考自http://blog.csdn.net/gemire/article/details/20832199