poj3255次短路算法

Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10539		Accepted: 3756

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

某街区有R条道路，N个路口，道路可以双向通过，问1路口到N路口的次短距离同一条路可以走多次

用最短路做法Dijkstra算法，dist[]记录最短路，dist2[]记录次短路

#include <iostream>
#include <vector>
#include <queue>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=5005;
const int inf=1e9;
int N,R;
struct edge
{
    int to,cost;
};
typedef pair<int ,int> P ;///first 是从1到second的最短路 second 是路口标号
vector<edge>G[maxn];///邻接表
int dist[maxn];///最短路
int dist2[maxn];///次短路
void solve()
{
    priority_queue<P ,vector<P>,greater<P> >que;
    fill(dist,dist+N,inf);
    fill(dist2,dist2+N,inf);
    dist[0]=0;
    //dist2[0]=0;
    que.push(P(0,0));
    while(!que.empty())
    {
        P p=que.top();///优先队列 ，用.top
        que.pop();
        int v=p.second ,d=p.first;
        if(dist2[v]<d)continue;
        for(int i=0; i<G[v].size(); i++)
        {
            edge e=G[v][i];
            int d2=d+e.cost;
            if(dist[e.to]>d2)
            {
                swap(dist[e.to],d2);
                que.push(P(dist[e.to],e.to));
            }
            if(dist2[e.to]>d2&&dist[e.to]<d2)
            {
                dist2[e.to]=d2;
                que.push(P(dist2[e.to],e.to));
            }
        }
    }
    printf("%d\n",dist2[N-1]);
}
int main()
{
    int from;
    while(cin>>N>>R)
    {
        edge now;
        for(int i=0;i<R;i++)
        {
            cin>>from>>now.to>>now.cost;
            from--;
            now.to--;///标号从  0-N-1！！
            G[from].push_back(now);
            swap(now.to,from);
            G[from].push_back(now);
        }
        solve();
    }
    return 0;
}