算法複習———dijkstra求次短路(poj3255)

題目：

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

 

題解：

　　次短路的模板題(注意是嚴格次短),詳細見http://www.javashuo.com/article/p-nltlftly-ck.html

　　但注意代碼中的註釋部分

代碼：

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<cctype>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
using namespace std;
priority_queue< pair<long long ,int> >que;
const int N=5005;
const int M=1e5+5;
const int inf=0x3f3f3f3f;
int sedis[N],dis[N],fst[N],nxt[M*2],go[M*2],val[M*2],tot,n,m;
inline int R(){
    char c;int f=0;
    for(c=getchar();c<'0'||c>'9';c=getchar());
    for(;c<='9'&&c>='0';c=getchar()) f=(f<<3)+(f<<1)+c-'0';
    return f;
}
inline void comb(int a,int b,int c){
    nxt[++tot]=fst[a],fst[a]=tot,go[tot]=b,val[tot]=c;
    nxt[++tot]=fst[b],fst[b]=tot,go[tot]=a,val[tot]=c;      
}
inline void getans(){
    memset(dis,inf,sizeof(dis));memset(sedis,inf,sizeof(sedis));
    dis[1]=0;que.push(make_pair(0,1));
    while(!que.empty()){
        int u=que.top().second,d=que.top().first;d=-d;que.pop();   //注意這裏的d!!!!!! 
        for(int e=fst[u];e;e=nxt[e]){
            int v=go[e];
            if(sedis[v]<val[e]+d) continue;
            else if(dis[v]>val[e]+d){
                sedis[v]=dis[v];
                dis[v]=val[e]+d;que.push(make_pair(-dis[v],v));
            }
            else if(dis[v]<val[e]+d&&sedis[v]>val[e]+d){
                sedis[v]=d+val[e];que.push(make_pair(-sedis[v],v));
            }
        }
    }
}
int main(){
    //freopen("a.in","r",stdin);
    n=R(),m=R();int a,b,c;
    for(int i=1;i<=m;i++){
        a=R(),b=R(),c=R();comb(a,b,c);
    }
    getans();cout<<sedis[n]<<endl;
    return 0;
}