poj 3255 次短路问题 Dijkstra 邻接表

Roadblocks
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Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14328		Accepted: 5037

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

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号路抠到N号路口的次短路经长度是多少？

次短路指的是币最短路长度长的次短路经。同一条边可以经过多次。

 
#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;
}