http://poj.org/problem?id=3255
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15680 | Accepted: 5510 |
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
mmp、、、无向图，双向边！！啊啊啊
1 #include <algorithm>
2 #include <cstdio>
3 #include <queue>
4
5 using namespace std;
6
7 const int INF(0x3f3f3f3f);
8 const int N(5000+105);
9 const int M(200000+5);
10
11 int hed[N],had[N],sumedge;
12 struct Edge
13 {
14 int v,next,w;
15 }edge1[M],edge2[M];
16 inline void ins(int u,int v,int w)
17 {
18 edge1[++sumedge].v=v;
19 edge1[sumedge].next=hed[u];
20 edge1[sumedge].w=w;
21 hed[u]=sumedge;
22 edge2[sumedge].v=u;
23 edge2[sumedge].next=had[v];
24 edge2[sumedge].w=w;
25 had[v]=sumedge;
26
27 edge1[++sumedge].v=u;
28 edge1[sumedge].next=hed[v];
29 edge1[sumedge].w=w;
30 hed[v]=sumedge;
31 edge2[sumedge].v=v;
32 edge2[sumedge].next=had[u];
33 edge2[sumedge].w=w;
34 had[u]=sumedge;
35 }
36
37 int dis[N];
38 bool inq[N];
39 void SPFA(int s)
40 {
41 for(int i=1;i<s;i++) dis[i]=INF;
42 dis[s]=0; inq[s]=1;
43 queue<int>que; que.push(s);
44 for(int u,v;!que.empty();)
45 {
46 u=que.front(); que.pop(); inq[u]=0;
47 for(int i=had[u];i;i=edge2[i].next)
48 {
49 v=edge2[i].v;
50 if(dis[v]>dis[u]+edge2[i].w)
51 {
52 dis[v]=dis[u]+edge2[i].w;
53 if(!inq[v]) que.push(v),inq[v]=1;
54 }
55 }
56 }
57 }
58
59 struct Node
60 {
61 int now,g;
62 bool operator < (Node a) const
63 {
64 return a.g+dis[a.now]<g+dis[now];
65 }
66 };
67 int Astar(int s,int t,int k)
68 {
69 priority_queue<Node>que;
70 int cnt=0; Node u,v;
71 u.g=0; u.now=s;
72 que.push(u);
73 for(;!que.empty();)
74 {
75 u=que.top(); que.pop();
76 if(u.now==t) cnt++;
77 if(cnt==k) return u.g;
78 for(int i=hed[u.now];i;i=edge1[i].next)
79 {
80 v.now=edge1[i].v;
81 v.g=u.g+edge1[i].w;
82 que.push(v);
83 }
84 }
85 return 0;
86 }
87
88 inline void read(int &x)
89 {
90 x=0; register char ch=getchar();
91 for(;ch>‘9‘||ch<‘0‘;) ch=getchar();
92 for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar()) x=x*10+ch-‘0‘;
93 }
94
95 int AC()
96 {
97 // freopen("block.in","r",stdin);
98 // freopen("block.out","w",stdout);
99
100 int n,m; read(n),read(m);
101 for(int v,u,w;m--;)
102 read(u),read(v),read(w),ins(u,v,w);
103 SPFA(n); printf("%d\n",Astar(1,n,2));
104 return 0;
105 }
106
107 int I_want_AC=AC();
108 int main(){;}
Astar AC
次短路正经做法：
SPFA跑出从1到i和从n到i的dis，枚举每条不在最短路上的边，更新ans
1 #include <algorithm>
2 #include <cstdio>
3 #include <queue>
4
5 using namespace std;
6
7 const int INF(0x3f3f3f3f);
8 const int N(5000+105);
9 const int M(100000+5);
10
11 int m,n,head[N],sumedge;
12 struct Edge
13 {
14 int v,next,w;
15 Edge(int v=0,int next=0,int w=0):
16 v(v),next(next),w(w){}
17 }edge[M<<1];
18 inline void ins(int u,int v,int w)
19 {
20 edge[++sumedge]=Edge(v,head[u],w);
21 head[u]=sumedge;
22 }
23
24 bool inq[N];
25 int d1[N],d2[N];
26 inline void SPFA(int s,int *dis)
27 {
28 for(int i=1;i<=n;i++)
29 inq[i]=0,dis[i]=INF;
30 dis[s]=0; inq[s]=1;
31 queue<int>que; que.push(s);
32 for(int u,v;!que.empty();)
33 {
34 u=que.front(); que.pop(); inq[u]=0;
35 for(int i=head[u];i;i=edge[i].next)
36 {
37 v=edge[i].v;
38 if(dis[v]>dis[u]+edge[i].w)
39 {
40 dis[v]=dis[u]+edge[i].w;
41 if(!inq[v]) que.push(v),inq[v]=1;
42 }
43 }
44 }
45 }
46
47 inline void read(int &x)
48 {
49 x=0; register char ch=getchar();
50 for(;ch>‘9‘||ch<‘0‘;) ch=getchar();
51 for(;ch>=‘0‘&&ch<=‘9‘;ch=getchar()) x=x*10+ch-‘0‘;
52 }
53
54 int AC()
55 {
56 freopen("block.in","r",stdin);
57 freopen("block.out","w",stdout);
58
59 read(n),read(m);
60 for(int v,u,w;m--;ins(u,v,w),ins(v,u,w))
61 read(u),read(v),read(w);
62 SPFA(1,d1); SPFA(n,d2);
63 int ans=INF,tmp;
64 for(int i,v,u=1;u<=n;u++)
65 {
66 for(int i=head[u];i;i=edge[i].next)
67 {
68 v=edge[i].v;
69 tmp=d1[u]+d2[v]+edge[i].w;
70 if(tmp>d1[n]&&ans>tmp) ans=tmp;
71 }
72 }
73 printf("%d\n",ans);
74 return 0;
75 }
76
77 int I_want_AC=AC();
78 int main(){;}
SPFA 跑次短路