poj 3255 求次大最短路

Roadblocks

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 5508		Accepted: 2088

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)

Source

USACO 2006 November Gold

题目大意：在一个图上有许多个农场，有个人从1农场出发，到他的朋友n农场去，他不想走一条最短路径，这次他想换条路走，要你帮他找一条次短路径，次短路的定义是，比最短路径长度短（可能有多条），但是不会比其他的路径长度长。而且告诉你数据中一定存在至少一条次短路。

思路 ：暴力每一条边 

求出从起点到达该边的起点的最短距离 求终点到达该边的终点距离 相加再加上边长  判断是否比最短路长 且最接近最短路

#include<stdio.h>
#include<queue>
#include<algorithm>
#include<vector>
#include<string.h>
using namespace std;
const int INF=10000000;
const int maxn=5002;
struct Edge{
    int from,to,dis;
    };
struct node{
    int d,u;
    bool operator <(const node &rhs) const {
    return d>rhs.d;//从小到大排列
    }
};
int mn;
struct Dij
{
    int n,m;//点数和边数
    vector<Edge>edges;//边列表
    vector<int>nd[maxn];//从每个节点出发的边编号 从0开始
    bool vis[maxn];//是否参观过
    int mmin[maxn];//s到各个点的最小距离
    void init(int n)
    {
        this->n=n;
        for(int i=0;i<n;i++)  nd[i].clear();//邻接表清空
        edges.clear();//清空边表
    }
 
    void addedge(int from,int to,int dis)
    //如果是无向图 每条无向边需要调用2次
    {
         edges.push_back((Edge){from,to,dis});//是{ 不是（
         m=edges.size();
         nd[from].push_back(m-1);
    }
 
    void dij(int s)//求s到所有点的距离
    {
        priority_queue<node>que;
        for(int i=0;i<n;i++) mmin[i]=INF;
        mmin[s]=0;
        memset(vis,0,sizeof(vis));
        que.push((node){0,s});//注意符号 不是括号
        while(!que.empty())
        {
            node x=que.top();que.pop();
            int u=x.u;
            if(vis[u]) continue;
            vis[u]=true;
            for(int i=0;i<nd[u].size();i++)
            {
                Edge& e=edges[nd[u][i]];
                int max1=0,max2=0;
                if(mmin[e.to]>mmin[u]+e.dis)
                 {
                   // printf("mmin[e.to]=%d\n",mmin[e.to]);
                   max1=mmin[e.to];
                   mmin[e.to]=mmin[u]+e.dis;
                   max2=mmin[e.to];
                   if(max1!=INF&&max2!=INF&&mn>max1-max2) mn=max1-max2;
                   que.push((node){mmin[e.to],e.to});
                 }
 
            }
        }
    }
};
 
vector<int>path;
int main()
{
    int n,m,x,y,z,s,e;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        mn=999999999;
        s=0;e=n-1;
        Dij solve[2];
        solve[0].init(n);
        solve[1].init(n);
           while(m--)
           {
                scanf("%d %d %d",&x,&y,&z);
                x--;y--;
                solve[0].addedge(x,y,z);solve[0].addedge(y,x,z);
                solve[1].addedge(x,y,z);solve[1].addedge(y,x,z);
           }
           solve[0].dij(s);//求1到所有点的距离
           solve[1].dij(e);//求n到所有点的距离
           int ans=INF;
            for(int i=0;i<solve[0].edges.size();i++)
            {
                Edge &edge=solve[0].edges[i];
                int st=edge.from,ed=edge.to;
                int mid=solve[0].mmin[st]+solve[1].mmin[ed]+edge.dis;
                if(mid>solve[0].mmin[e])
                {
                    if(ans>mid) ans=mid;
                }
            }
            printf("%d\n",ans);
     }
     return 0;
}

参考：

http://blog.csdn.net/wangjian8006/article/details/7992280