POJ 3255 Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15645		Accepted: 5504

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 intersectionN.

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)

Source

USACO 2006 November Gold

 

题意：求次短路

模板

丫的无向边

屠龙宝刀点击就送

1 A*

堆优化dijkstra + 超级读入优化+输出优化+黑科技。

COGS 上秒杀所有

稳居Rank 1 蛤蛤

#include <ctype.h>
#include <cstdio>
#include <vector>
#include <queue>
#define BUF 100000010
#define M 100005
#define N 5005
char Buf[BUF],*buf=Buf;
struct Edge
{
    int next,to,w;
};
Edge edge1[M<<1],edge2[M<<1];
using namespace std;
int n,r,cnt,dis[N],head1[N],head2[N];
bool vis[N];
struct node
{
    int to,g;
    bool operator<(node a)const
    {
        return g+dis[to]>a.g+dis[a.to];
    }
};
struct Node
{
    int x,y;
    bool operator<(Node a)const
    {
        return y>a.y;
    }
};
inline void Read(register int &x)
{
    for(x=0;!isdigit(*buf);++buf);
    for(;isdigit(*buf);x=x*10+*buf-'0',++buf);
}
inline void ins(int u,int v,int w)
{
    edge1[++cnt].next=head1[u];
    edge1[cnt].to=v;
    edge1[cnt].w=w;
    head1[u]=cnt;
    edge2[cnt].next=head2[v];
    edge2[cnt].to=u;
    edge2[cnt].w=w;
    head2[v]=cnt;
    edge1[++cnt].next=head1[v];
    edge1[cnt].to=u;
    edge1[cnt].w=w;
    head1[v]=cnt;
    edge2[cnt].next=head2[u];
    edge2[cnt].to=v;
    edge2[cnt].w=w;
    head2[u]=cnt;
}
void dij(int s)
{
    for(int i=1;i<n;++i) dis[i]=0x7fffffff;
    priority_queue<Node>q;
    dis[s]=0;
    Node a;
    a.x=s;
    a.y=dis[s];
    q.push(a);
    for(Node now;!q.empty();)
    {
        now=q.top();
        q.pop();
        if(vis[now.x]) continue;
        vis[now.x]=1;
        for(register int i=head2[now.x];i;i=edge2[i].next)
        {
            int v=edge2[i].to;
            if(dis[v]>dis[now.x]+edge2[i].w)
            {
                dis[v]=dis[now.x]+edge2[i].w;
                Node tmp;
                tmp.x=v;
                tmp.y=dis[v];
                q.push(tmp);
            }
        }
    }
}
int Astar(register int s,register int t)
{
    priority_queue<node>q;
    node a;
    a.to=s;a.g=0;
    q.push(a);
    register int ct=0;
    for(node now;!q.empty();)
    {
        now=q.top();q.pop(); 
        if(now.to==t) ct++;
        if(ct==2) return now.g;
        for(register int i=head1[now.to];i;i=edge1[i].next)
        {
            int v=edge1[i].to;
            node tmp;
            tmp.to=v;tmp.g=now.g+edge1[i].w;
            q.push(tmp);
        }
    }
    return 0;
}
void write(register int x)
{
    if(x>9) write(x/10);
    putchar(x%10+'0');
}
int Main()
{
#define MINE
#ifdef MINE
    freopen("block.in","r",stdin);
    freopen("block.out","w",stdout);
    fread(buf,1,BUF,stdin);
#endif
    Read(n);
    Read(r);
    for(register int x,y,z;r--;)
    {
        Read(x);
        Read(y);
        Read(z);
        ins(x,y,z);
    }
    dij(n);
    write(Astar(1,n));
    return 0;
}
int sb=Main();
int main(int argc,char *argv[]) {;}

2. spfa

#include <ctype.h>
#include <cstdio>
#include <vector>
#include <queue>
#define M 100005
#define N 5005

struct Edge
{
    int next,from,to,w;
    Edge (int next=0,int from=0,int to=0,int w=0) :next(next),from(from),to(to),w(w) {}
}edge[M<<1];
using namespace std;
int n,r,cnt,dis1[N],dis2[N],head[N];
bool vis[N];
inline void Read(int &x)
{
    bool f=0;
    register char ch=getchar();
    for(x=0;!isdigit(ch);ch=getchar()) if(ch=='-') f=1;
    for(;isdigit(ch);x=x*10+ch-'0',ch=getchar());
    x=f?-x:x;
}
void ins(int u,int v,int w)
{
    edge[++cnt]=Edge(head[u],u,v,w);
    head[u]=cnt;
    edge[++cnt]=Edge(head[v],v,u,w);
    head[v]=cnt;
}
void spfa(int s,int *dis)
{
    queue<int>q;
    q.push(s);
    for(int i=1;i<=n;++i) dis[i]=0x7fffffff;
    dis[s]=0;
    vis[s]=1;
    for(int now;!q.empty();)
    {
        now=q.front();
        q.pop();
        vis[now]=0;
        for(int i=head[now];i;i=edge[i].next)
        {
            int v=edge[i].to;
            if(dis[v]>dis[now]+edge[i].w)
            {
                dis[v]=dis[now]+edge[i].w;
                if(!vis[v])
                {
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}
int main()
{
    Read(n);
    Read(r);
    for(int x,y,z;r--;)
    {
        Read(x);
        Read(y);
        Read(z);
        ins(x,y,z);
    }
    spfa(1,dis1);
    spfa(n,dis2);
    int Min=dis1[n],ans=0x7fffffff,dist;
    for(int i=1;i<=cnt;++i)
    {
        int u=edge[i].from,v=edge[i].to;
        dist=dis1[u]+edge[i].w+dis2[v];
        if(dist>Min&&dist<ans) ans=dist;
    }
    printf("%d\n",ans);
    return 0;
}