POJ 3255 Roadblocks （次短路模板）

Roadblocks

http://poj.org/problem?id=3255

Time Limit: 2000MS		Memory Limit: 65536K

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)

 

题意：求严格次短路

 

次短路一定是由最短路上出去经过其他的边在回到最短路上

所以先求出每个点到1的最短路dis1，每个点到n的最短路dis2

枚举每一条边

那么经过这条边u-->v的最短路为dis1[u]+w+dis2[v]

如果它大于最短路，更新答案

 

#include<queue>
#include<cstdio>
#include<cstring>
#define N 5001
#define M 100001
using namespace std;
int front[N],to[M<<],nxt[M<<],val[M<<],from[M<<],tot;
int dis1[N],dis2[N];
bool vis[N];
queue<int>q;
void add(int u,int v,int w)
{
    to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; val[tot]=w; from[tot]=u;
    to[++tot]=u; nxt[tot]=front[v]; front[v]=tot; val[tot]=w; from[tot]=v;
}
void spfa(int s,int *dis)
{
    //memset(dis,63,sizeof(dis));
    memset(vis,false,sizeof(vis));
    q.push(s);
    vis[s]=true;
    dis[s]=;
    int now;
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        vis[now]=false;
        for(int i=front[now];i;i=nxt[i])
            if(dis[to[i]]>dis[now]+val[i])
            {
                dis[to[i]]=dis[now]+val[i];
                if(!vis[to[i]])
                {
                    vis[to[i]]=true;
                    q.push(to[i]);
                }
            }
    }
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    int u,v,w;
    for(int i=;i<=m;i++)
    {
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w);
    }
    memset(dis1,,sizeof(dis1));
    memset(dis2,,sizeof(dis2));
    spfa(,dis1);
    spfa(n,dis2);
    int minn=dis1[n],tmp,ans=0x7fffffff;
    for(int i=;i<=tot;i++)
    {
        u=from[i]; v=to[i];
        tmp=dis1[u]+val[i]+dis2[v];
        if(tmp>minn && tmp<ans) ans=tmp;
    }
    printf("%d",ans);
}