POJ3255:求解[次短路]->Dijkstra

Roadblocks
Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 28456		Accepted: 9683
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)

#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

#define MAXN 5005
#define MAXR 100005
#define INF (1 << 30)
typedef pair<int, int> P;

struct edge
{
    /* data */
    int to, cost;
    edge(int t, int c)
    {
        to = t;
        cost = c;
    }
};

vector<edge> G[MAXN];

int dist[MAXN];  //最短距离
int dist2[MAXN]; //次短距离

int n, r;

int main()
{
    scanf("%d%d", &n, &r);

    int a, b, d;
    for (int i = 1; i <= r; ++i)
    {
        scanf("%d%d%d", &a, &b, &d);
        G[a].push_back(edge(b, d));
        G[b].push_back(edge(a, d));
    }

    priority_queue<P, vector<P>, greater<P>> q;
    fill(dist, dist + n + 1, INF);
    fill(dist2, dist2 + n + 1, INF);

    dist[0] = 0;
    dist[1] = 0;
    q.push(P(0, 1));

    while (!q.empty())
    {
        P p = q.top();
        q.pop();
        int v = p.second;
        int 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;
            /**
             * 到e.to的次短路有两种情况
                * 是起点到v的最短路+v到e.to的边 
                * 是起点到v的次短路+v到e.to的边
            */
            if (dist[e.to] > d2)
            {
                //更新到e.to的最短路,这里使用swap的目的是，把原来的最短路更新,并使其在下一个if中，能够成接下来的次短路。
                swap(dist[e.to], d2);
                q.push(P(dist[e.to], e.to));
            }
            if (dist2[e.to] > d2 && dist[e.to] < d2) //在d2不是最短路，且当前次短路不是最优的情况下，更新次短路
            {
                dist2[e.to] = d2;
                q.push(P(dist2[e.to], e.to));
            }
        }
    }

    printf("%d\n", dist2[n]);
}