[POJ3255] 地砖RoadBlocks

[POJ3255] 地砖RoadBlocks

 

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)

Source

USACO 2006 November Gold

 

求次短路 

可以用A*

这里我用的dijs+堆优化

题目有点坑 

这个可以来回跑。。 从1->3->1->...->n;的路径合法

SPFA好像没法重复跑 

所以我WA了又WA 就改了dijs

 

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <ctype.h>
 4 #include <cstdio>
 5 #include <vector>
 6 #include <queue>
 7 
 8 using namespace std;
 9 
10 const int MAXN=5010;
11 const int INF=1e9;
12 
13 int n,m;
14 
15 int dis[MAXN],dis2[MAXN];
16 
17 typedef pair<int,int> P;
18 
19 struct edge {
20     int to;
21     int val;
22 };
23 vector<edge> G[MAXN];
24 
25 inline void read(int&x) {
26     int f=1;register char c=getchar();
27     for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar());
28     for(;isdigit(c);x=x*10+c-48,c=getchar());
29     x=x*f;
30 }
31 
32 inline void SPFA() {
33     priority_queue<P,vector<P>,greater<P> >Q;
34     for(int i=0;i<=n;++i) dis[i]=dis2[i]=INF;
35     Q.push(P(1,0));
36     dis[1]=0;
37     while(!Q.empty()) {
38         P u=Q.top();
39         Q.pop();
40         int v=u.first,V=u.second;
41         if(dis2[v]<V) continue;
42         for(int i=0;i<G[v].size();++i) {
43             edge e=G[v][i];
44             int d=V+e.val;
45             if(dis[e.to]>d) {
46                 swap(dis[e.to],d);
47                 Q.push(P(e.to,dis[e.to]));
48             }
49             if(dis2[e.to]>d&&d>dis[e.to]) {
50                 dis2[e.to]=d;
51                 Q.push(P(e.to,dis2[e.to]));
52             }
53         }
54     }
55     return;
56 }
57 
58 int hh() {
59 //    freopen("1.in","r",stdin);
60 //    freopen("2.out","w",stdout);
61     read(n);read(m);
62     for(int x;m--;) {
63         edge p;
64         read(x);read(p.to);read(p.val);
65 //        --x,--p.to;
66         G[x].push_back(p);
67         swap(p.to,x);
68         G[x].push_back(p);
69     }
70     SPFA();
71     printf("%d
",dis2[n]);
72     return 0;
73 }
74 
75 int sb=hh();
76 int main() {;}

#include <ctype.h>
#include <cstdio>
#include <queue>

const int MAXN=5010;
const int MAXM=100010;
const int INF=0x3f3f3f3f;

int n,R,last,ans;

int dis[MAXN];

bool vis[MAXN];

struct SKT {
    int v,dist;
    bool operator < (SKT b) const{
        return dist+dis[v]>b.dist+dis[b.v];
    }
};
SKT s;

struct edge {
    int to;
    int val;
    int next;
    edge() {}
    edge(int to,int val,int next):to(to),val(val),next(next) {}
};
edge e[MAXM<<1];

int head[MAXN],tot;

inline void read(int&x) {
    int f=1;register char c=getchar();
    for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar());
    for(;isdigit(c);x=x*10+c-48,c=getchar());
    x=x*f;
}

inline void add(int x,int y,int v) {
    e[++tot]=edge(y,v,head[x]);head[x]=tot;
    e[++tot]=edge(x,v,head[y]);head[y]=tot;
}

void SPFA() {
    std::queue<int> Q;
    for(int i=1;i<=n;++i) dis[i]=INF,vis[i]=false;
    dis[n]=0;
    vis[n]=true;
    Q.push(n);
    while(!Q.empty()) {
        int now=Q.front();
        Q.pop();
        vis[now]=false;
        for(int i=head[now];i;i=e[i].next) {
            int u=e[i].to;
            if(dis[u]>dis[now]+e[i].val) {
                dis[u]=dis[now]+e[i].val;
                if(!vis[u]) Q.push(u),vis[u]=true;
            }
        }
    }
    return;
}

void Astar() {
    std::priority_queue<SKT> Q;
    s.v=1;s.dist=0;
    Q.push(s);
    while(!Q.empty()) {
        SKT now=Q.top();
        Q.pop();
        if(now.v==n) ++ans,last=now.dist;
        if(ans==2) {
            printf("%d
",now.dist);
            return;
        }
        for(int i=head[now.v];i;i=e[i].next) {
            int u=e[i].to;
            SKT p=now;
            p.v=u;
            p.dist+=e[i].val;
            Q.push(p);
        }
    }
    return;
}

int hh() {
    freopen("block.in","r",stdin);
    freopen("block.out","w",stdout);
    read(n);read(R);
    for(int x,y,z;R--;) {
        read(x);read(y);read(z);
        add(x,y,z);
    }
    SPFA();
    Astar();
    return 0;
}

int sb=hh();
int main() {;}