UVA10917 Walk Through the Forest

Posted 2020-10-22 嘒彼小星

题目大意：Jimmy下班后决定每天沿着一条不同的路径回家，欣赏不同的风景。他打算只沿着满足如下条件的（A,B）道路走：存在一条从B出发回家的路，比所有从A出发回家的路径都短。你的任务是计算一共有多少条不同的回家路径。其中公司的编号为1，家的编号为2.

 

题解：算出每个点到2的最短路，对于每条边(i,j)，“存在一条从B出发回家的路，比所有从A出发回家的路径都短”，即d[B] < d[A]，说明能从A走到B，于是建新图A->B。不难发现是一个DAG（因为有环就会出现逻辑矛盾）。建反图dp即可

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 #include <map>
 9 #include <string> 
10 #include <cmath> 
11 #define min(a, b) ((a) < (b) ? (a) : (b))
12 #define max(a, b) ((a) > (b) ? (a) : (b))
13 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
14 template<class T>
15 inline void swap(T &a, T &b)
16 {
17     T tmp = a;a = b;b = tmp;
18 }
19 inline void read(int &x)
20 {
21     x = 0;char ch = getchar(), c = ch;
22     while(ch < ‘0‘ || ch > ‘9‘) c = ch, ch = getchar();
23     while(ch <= ‘9‘ && ch >= ‘0‘) x = x * 10 + ch - ‘0‘, ch = getchar();
24     if(c == ‘-‘) x = -x;
25 }
26 const int INF = 0x3f3f3f3f;
27 const int MAXN = 1000 + 10;
28 struct Edge
29 {
30     int u,v,w,nxt;
31     Edge(int _u, int _v, int _w, int _nxt){u = _u;v = _v;w = _w;nxt = _nxt;}
32     Edge(){}
33 }edge1[MAXN * MAXN], edge2[MAXN * MAXN];
34 int head1[MAXN], cnt1, head2[MAXN], cnt2;
35 inline void insert1(int a, int b, int c){edge1[++cnt1] = Edge(a,b,c,head1[a]), head1[a] = cnt1;}
36 inline void insert2(int a, int b, int c){edge2[++cnt2] = Edge(a,b,c,head2[a]), head2[a] = cnt2;}
37 struct Node
38 {
39     int u,w;
40     Node(int _u, int _w){u = _u;w = _w;}
41     Node(){}
42 };
43 struct cmp
44 {
45     bool operator()(Node a, Node b)
46     {
47         return a.w > b.w;
48     }
49 };
50 std::priority_queue<Node, std::vector<Node>, cmp> q;
51 int n,m,d[MAXN],vis[MAXN],tmp1,tmp2,tmp3,dp[MAXN];
52 void dij(int S)
53 {
54     memset(d, 0x3f, sizeof(d)), d[S] = 0, memset(vis, 0, sizeof(vis)), q.push(Node(S, 0));
55     while(q.size())
56     {
57         Node now = q.top();q.pop();
58         if(vis[now.u]) continue;vis[now.u] = 1;
59         for(int pos = head1[now.u];pos;pos = edge1[pos].nxt)
60         {
61             int v = edge1[pos].v;
62             if(vis[v]) continue;
63             if(d[v] > d[now.u] + edge1[pos].w) d[v] = d[now.u] + edge1[pos].w, q.push(Node(v, d[v]));
64         }
65     }
66 }
67 int Dp(int x)
68 {
69     if(vis[x]) return dp[x];
70     for(int pos = head2[x];pos;pos = edge2[pos].nxt)
71     {
72         int v = edge2[pos].v;
73         dp[x] += Dp(v);
74     }
75     vis[x] = 1;
76     return dp[x];
77 }
78 int main()
79 {
80     while(scanf("%d", &n) != EOF && n)
81     {
82         read(m), memset(head1, 0, sizeof(head1)), memset(head2, 0, sizeof(head2)), cnt1 = 0, cnt2 = 0;
83         for(int i = 1;i <= m;++ i) read(tmp1), read(tmp2), read(tmp3), insert1(tmp1, tmp2, tmp3), insert1(tmp2, tmp1, tmp3);
84         dij(2);
85         for(int i = 1;i <= cnt1;++ i)
86             if(d[edge1[i].v] < d[edge1[i].u])
87                 insert2(edge1[i].v, edge1[i].u, edge1[i].w);
88         memset(vis, 0, sizeof(vis)), memset(dp, 0, sizeof(dp)), vis[1] = 0, dp[1] = 1;
89         printf("%d\n", Dp(2)); 
90     }
91     return 0;
92 }

UVA10917