A Walk Through the Forest
Time Limit:1000MS Memory Limit:65536K
Total Submit:48 Accepted:15
Description
Jimmy experiences a lot of stress at work these days, especially since his
accident made working difficult. To relax after a hard day, he likes to walk
home. To make things even nicer, his office is on one side of a forest, and his
house is on the other. A nice walk through the forest, seeing the birds and
chipmunks is quite enjoyable.
The forest is beautiful, and Jimmy wants to take a different route everyday. He
also wants to get home before dark, so he always takes a path to make progress
towards his house. He considers taking a path from A to B to be progress if
there exists a route from B to his home that is shorter than any possible route
from A. Calculate how many different routes through the forest Jimmy might
take.
Input
Input contains several test cases followed by a line containing 0. Jimmy
has numbered each intersection or joining of paths starting with 1. His office
is numbered 1, and his house is numbered 2. The first line of each test case
gives the number of intersections N, 1 < N ≤ 1000, and the number of paths
M. The following M lines each contain a pair of intersections a b and an
integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between
intersection a and a different intersection b. Jimmy may walk a path any
direction he chooses. There is at most one path between any pair of
intersections.
Output
For each test case, output a single integer indicating the number of
different routes through the forest. You may assume that this number does not
exceed 2147483647.
Sample
Input
5 6
1 3 2
1 4 2
3 4 3
1 5 12
4 2 34
5 2 24
7 8
1 3 1
1 4 1
3 7 1
7 4 1
7 5 1
6 7 1
5 2 1
6 2 1
0
Sample
Output
2
4
【思路】
最短路+记忆化搜索。
SPFA预处理出每个点到达home的最短距离d，然后沿着d变小的边记忆化搜索路径条数。
【代码】
1 #include<cstdio>
2 #include<queue>
3 #include<vector>
4 #include<cstring>
5 using namespace std;
6
7 const int maxn = 1000+10;
8 const int INF=1<<30;
9 struct Edge{
10 int u,v,w,next;
11 }e[2*maxn*maxn];
12 int en,front[maxn];
13
14 int n,m;
15
16 inline void AddEdge(int u,int v,int w) {
17 en++; e[en].v=v; e[en].w=w; e[en].next=front[u]; front[u]=en;
18 }
19
20 int d[maxn];
21 void SPFA(int s) {
22 int inq[maxn];
23 queue<int> q;
24 memset(inq,0,sizeof(inq));
25 for(int i=1;i<=n;i++) d[i]=INF;
26
27 d[s]=0; inq[s]=1; q.push(s);
28 while(!q.empty()) {
29 int u=q.front(); q.pop(); inq[u]=0;
30 for(int i=front[u];i>=0;i=e[i].next) {
31 int v=e[i].v,w=e[i].w;
32 if(d[v]>d[u]+w) {
33 d[v]=d[u]+w;
34 if(!inq[v]) {
35 inq[v]=1;
36 q.push(v);
37 }
38 }
39 }
40 }
41 }
42
43 int f[maxn];
44 int dp(int u) {
45 int& ans=f[u];
46 if(ans>=0) return ans;
47
48 if(u==2) return ans=1;
49 ans=0;
50 for(int i=front[u];i>=0;i=e[i].next) {
51 int v=e[i].v;
52 if(d[v]<d[u]) ans += dp(v); //只沿着d更小的走
53 }
54 return ans;
55 }
56 int main() {
57 while(scanf("%d%d",&n,&m)==2) {
58 en=-1;
59 memset(front,-1,sizeof(front));
60 int u,v,w;
61 for(int i=0;i<m;i++) {
62 scanf("%d%d%d",&u,&v,&w);
63 AddEdge(u,v,w);
64 AddEdge(v,u,w);
65 }
66 SPFA(2);
67 memset(f,-1,sizeof(f));
68 printf("%d\n",dp(1));
69 }
70 return 0;
71 }