POJ 3255 Roadblocks 次短路

和Dijksta求最短路一样，只是要维护两个数组：最短路d1,次短路d2。然后更新的时候注意细节。

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<cmath>
#include<climits>
#include<string>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define pb(a) push(a)
#define INF 0x1f1f1f1f
#define lson idx<<1,l,mid
#define rson idx<<1|1,mid+1,r
#define PI  3.1415926535898
template<class T> T min(const T& a,const T& b,const T& c) {
    return min(min(a,b),min(a,c));
}
template<class T> T max(const T& a,const T& b,const T& c) {
    return max(max(a,b),max(a,c));
}
void debug() {
#ifdef ONLINE_JUDGE
#else

    freopen("in.txt","r",stdin);
    //freopen("d:\\out1.txt","w",stdout);
#endif
}
int getch() {
    int ch;
    while((ch=getchar())!=EOF) {
        if(ch!=' '&&ch!='\n')return ch;
    }
    return EOF;
}

struct HeapNode
{
    int d,u;
    bool operator < (const HeapNode &ant) const
    {
        return ant.d<d;
    }
};

struct Edge
{
    int from,to,dist;
};

const int maxn=5005;

vector<int> g[maxn];
vector<Edge> edge;
int n;
int d1[maxn],d2[maxn];

void init()
{
    for(int i=1;i<=n;i++)
        g[i].clear();
    edge.clear();
}

void add(int u,int v,int w)
{
    Edge e=(Edge){u,v,w};
    edge.push_back(e);
    g[u].push_back(edge.size()-1);
}
void solve(int s)
{
    for(int i=1;i<=n;i++)
        d1[i]=d2[i]=INF;
    priority_queue<HeapNode> q;
    d1[s]=0;
    q.push((HeapNode){0,s});
    while(!q.empty())
    {
        HeapNode x=q.top(); q.pop();
        if(x.d>d2[x.u])continue;
        int u=x.u;
        for(int i=0;i<g[u].size();i++)
        {
            Edge &e=edge[g[u][i]];
            int v=e.to;
            int d=x.d;
            if(e.dist+d<d1[v])
            {
                d2[v]=d1[v];
                d1[v]=d+e.dist;
                q.push((HeapNode){d1[v],v});
            }else if(e.dist+d<d2[v]&&e.dist+d!=d1[v])
            {
                d2[v]=d+e.dist;
                q.push((HeapNode){d2[v],v});
            }
        }
    }
}
int main()
{
    int m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        for(int i=1;i<=m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }

        solve(1);
        printf("%d\n",d2[n]);
    }
    return 0;
}