[第2短路变形]LightOJ 1099 - Not the Best

1099 - Not the Best

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Time Limit:2 second(s)

Memory Limit:32 MB

Robin has moved to a small village and sometimes enjoys returning to visit one of his best friends. He does not want to get to his old home too quickly, because he likes the scenery along the way. He has decided to take the second-shortest rather than the shortest path. He knows there must be some second-shortest path.

The countryside consists ofRbidirectional roads, each linking two of theNintersections, conveniently numbered from1toN. Robin starts at intersection1, and his 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

Input starts with an integerT (≤ 10), denoting the number of test cases.

Each case contains two integersN (1 ≤ N ≤ 5000)andR (1 ≤ R ≤ 10⁵). Each of the nextRlines contains three space-separated integers:u, vandwthat describe a road that connects intersectionsuandvand has lengthw (1 ≤ w ≤ 5000).

Output

For each case, print the case number and the second best shortest path as described above.

Sample Input	Output for Sample Input
2 3 3 1 2 100 2 3 200 1 3 50 4 4 1 2 100 2 4 200 2 3 250 3 4 100	Case 1: 150 Case 2: 450

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PROBLEM SETTER: JANE ALAM JAN

传送门：http://lightoj.com/volume_showproblem.php?problem=1099

题意：对第2短路做了一个新的定义，如果存在多条最短路，那么一条比最短路长，而且比其他所有路径短的路径才是第2短路。

思路：和上一题差不多，直接用A*，然后改变退出条件即可，即 cur.f > dist[src]

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
const int MAXN = 5555,MAXM = 111111;
int inQ[MAXN],dist[MAXN],t,n,m,head[MAXN],ecnt;
struct edge{
    int v,w,nxt;
    edge(){nxt=-1;}
    edge(int vv,int ww,int nx):v(vv),w(ww),nxt(nx){}
}next[MAXM<<1];
void init(){
    memset(inQ,0,sizeof(inQ));
    fill(dist,dist+MAXN,0x3fffffff);
    memset(head,-1,sizeof(head));
    ecnt = 0;
}
void add(int u,int v,int w){
    next[ecnt++] = edge(v,w,-1);
    next[ecnt++] = edge(u,w,-1);
    next[ecnt-2].nxt = head[u];
    head[u] = ecnt-2;
    next[ecnt-1].nxt = head[v];
    head[v] = ecnt-1;
}
struct node{
    int f,g,v;
    node(){}
    node(int ff,int gg,int vv):f(ff),g(gg),v(vv){}
    bool operator < (const node &n) const{
        return n.f < f;
    }
};
void spfa(int src){
    queue<int> Q;
    Q.push(src);
    inQ[src] = 1;
    dist[src] = 0;
    while(!Q.empty()){
        int now = Q.front();
        Q.pop();inQ[now]=0;
        for(int i=head[now];i!=-1;i=next[i].nxt){
            if(dist[next[i].v]>dist[now]+next[i].w){
                dist[next[i].v] = dist[now]+next[i].w;
                if(!inQ[next[i].v]){
                    inQ[next[i].v]=1;
                    Q.push(next[i].v);
                }
            }
        }
    }
}
int Astar(int src){
    priority_queue<node> PQ;
    memset(inQ,0,sizeof(inQ));
    PQ.push(node(dist[src],0,src));
    while(!PQ.empty()){
        node cur = PQ.top();
        PQ.pop();inQ[cur.v]++;
        //cout<<cur.f<<endl;
        if(cur.f>dist[src]){
            return cur.f;
        }
        for(int i=head[cur.v];i!=-1;i=next[i].nxt){
            node expand(cur.g+dist[next[i].v]+next[i].w,cur.g+next[i].w,next[i].v);
            PQ.push(expand);
        }
    }
    return -1;
}
int main(){
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++){
        scanf("%d%d",&n,&m);
        init();
        while(m--){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);a--,b--;
            add(a,b,c);
        }
        spfa(n-1);
        printf("Case %d: %d\n",cas,Astar(0));
    }
    return 0;
}