POJ 3255 Roadblocks(次短路，Dijkstra变形+邻接表存储)

POJ 3255 Roadblocks

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)

这道题的N最大取到5000，用邻接矩阵做会直接Memory Limit Exceed，所以要用邻接表来做。

总结一下两种常用的邻接表写法：

1.利用vector的模拟链表

#include<vector>
using namespace std;
typedef struct{
    int to,w;
}node;
vector<node>map[max];

2.静态建表(链式前向星)

据说是目前建图和遍历效率最高的存储方式。

head数组初始化为-1即可。

typedef struct{
    int to,w,next;
}node;
node edge[M];
int head[N];

存储过程如下:

edge[top].to=e;
edge[top].w=w;
edge[top].next=head[s];
head[s]=top++;

关于题目还有一些细节和松弛过程变形的思路还需要理理，稍后补。

（因为照着别人的代码改了一个地方后AC了，还不清楚为什么= =）

AC代码：

#include<cstdio>
#include<cstring>
#define INF 0x3f3f3f
int N,R,top;
typedef struct{
    int to,w,next;
}edgenode;
edgenode edge[100010*2];
int head[5050];
bool vis[5050][2];
int d[5050];
int f[5050];
int add(int s,int e,int w){
    edge[top].to=e;
    edge[top].w=w;
    edge[top].next=head[s];
    head[s]=top++;
    return 0;
}
int main(){
    int i,j,k,m,n;
    int s,e,w,mins,pics,py;
    top=0;
    scanf("%d %d",&N,&R);
    for(i=1;i<=N;i++){
        d[i]=f[i]=INF;
        head[i]=-1;
        vis[i][0]=vis[i][1]=false;
    }
    for(i=1;i<=R;i++){
        scanf("%d %d %d",&s,&e,&w);
        add(s,e,w);
        add(e,s,w);
    }
    d[1]=0;
    for(i=1;i<=2*N;i++){
        mins=INF;
        for(j=1;j<=N;j++){
            if(!vis[j][0]&&d[j]<mins){
                pics=j;
                py=0;
                mins=d[j];
            }
            else if(!vis[j][1]&&f[j]<mins){
                pics=j;
                py=1;
                mins=f[j];
            }
        }
        if(mins==INF)
	    break;
        vis[pics][py]=true;
        for(k=head[pics];k!=-1;k=edge[k].next){
            m=edge[k].w;
            n=edge[k].to;
            if(d[n]>(mins+m)){
                f[n]=d[n];
                d[n]=mins+m;
            }
            else if(f[n]>(mins+m))
                f[n]=mins+m;
        }
    }
    printf("%d\n",f[N]);
    return 0;
}