POJ 3255 Roadblocks

求次短路径。先求出从1出发的单源最短路，用d[0][i]表示，再求出从N出发的单源最短路，

用d[1][i]表示。次短路应该是在d[0][a] +　d[1][b] + w[a][b]中选，先用heap+Dijktra做两遍最短路，然后再求次短路即可。

 

/*Accepted    2792K    110MS    C++    2128B    2012-09-22 10:29:43*/

#include <stdio.h>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

#include <queue>

using namespace std;



const int MAXN = 5050, MAXR = 100100, INF = 0x3f3f3f3f;

int first[MAXN], next[MAXR * 2], v[MAXR * 2], w[MAXR * 2];

int d[2][MAXN];

int N, R, e, ans;

bool vis[MAXN];

typedef pair<int, int> pii;



void addedge(int a, int b, int c)

{

    /*

    for(int i = first[a]; i != -1; i = next[i])

    {

        if(v[i] == b)

        {

            if(w[i] > c) w[i] = c;

            return;

        }

    }//判重边

    */

    next[e] = first[a], v[e] = b, w[e] = c;

    first[a] = e ++;

}



void ReadGraph()

{

    int a, b, c;

    memset(first, -1, sizeof first);

    e = 0;

    while(R --)

    {

        scanf("%d%d%d", &a, &b, &c);

        addedge(a, b, c);

        addedge(b, a, c);

    }

}



void Dijkstra(int src, int t)

{

    priority_queue<pii, vector<pii>, greater<pii> > q;

    for(int i = 0; i <= N; i ++)

        d[t][i] = INF;

    d[t][src] = 0;

    q.push(make_pair(d[t][src], src));

    memset(vis, false, sizeof vis);

    while(!q.empty())

    {

        pii u = q.top();

        q.pop();

        int x = u.second;

        //if(d[t][x] != u.first) continue;

        if(vis[x]) continue;

        vis[x] = true;

        for(int k = first[x]; k != -1; k = next[k])

        {

            if(d[t][v[k]] > d[t][x] + w[k])

            {

                d[t][v[k]] = d[t][x] + w[k];

                q.push(make_pair(d[t][v[k]], v[k]));

            }

        }

    }

}



void cal()

{

    Dijkstra(1, 0); //到1的最短路

    Dijkstra(N, 1); //到N的最短路

    int shortest = d[0][N];

    ans = INF;

    for(int i = 0; i < e; i += 2)

    {

        int e1 = i, e2 = i + 1;

        int a = v[e2], b = v[e1];

        int res = d[0][a] + d[1][b] + w[e1];

        if (res < ans && res > shortest)

            ans = res;

        res = d[1][a] + d[0][b] + w[e2];

        if (res < ans && res > shortest)

            ans = res;

    }

}



int main()

{

    while(scanf("%d%d", &N, &R) != EOF)

    {

        ReadGraph();

        cal();

        printf("%d\n", ans);

    }

    return 0;

}