codeforces 144 D Missile Silos(最短)
19381 ワード
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Missile Silos
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that: between any two cities no more than one road exists; each road connects two different cities; from each city there is at least one way to any other city by the roads.
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Sample test(s)
Input
Output
Input
Output
Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
タイトル:
図を1枚与えて、図の上のすべてのs点までの距離がdのいくつかの点があることを聞きます.
一度最短絡し,sからすべての点までの最短距離を得た.次に、各エッジを列挙し、エッジに要求を満たす点があるかどうかを統計します.
コード君
Missile Silos
A country called Berland consists of n cities, numbered with integer numbers from 1 to n. Some of them are connected by bidirectional roads. Each road has some length. There is a path from each city to any other one by these roads. According to some Super Duper Documents, Berland is protected by the Super Duper Missiles. The exact position of the Super Duper Secret Missile Silos is kept secret but Bob managed to get hold of the information. That information says that all silos are located exactly at a distance l from the capital. The capital is located in the city with number s.
The documents give the formal definition: the Super Duper Secret Missile Silo is located at some place (which is either city or a point on a road) if and only if the shortest distance from this place to the capital along the roads of the country equals exactly l.
Bob wants to know how many missile silos are located in Berland to sell the information then to enemy spies. Help Bob.
Input
The first line contains three integers n, m and s (2 ≤ n ≤ 105, 1 ≤ s ≤ n) — the number of cities, the number of roads in the country and the number of the capital, correspondingly. Capital is the city no. s.
Then m lines contain the descriptions of roads. Each of them is described by three integers vi, ui, wi (1 ≤ vi, ui ≤ n, vi ≠ ui, 1 ≤ wi ≤ 1000), where vi, ui are numbers of the cities connected by this road and wi is its length. The last input line contains integer l (0 ≤ l ≤ 109) — the distance from the capital to the missile silos. It is guaranteed that:
Output
Print the single number — the number of Super Duper Secret Missile Silos that are located in Berland.
Sample test(s)
Input
4 6 1
1 2 1
1 3 3
2 3 1
2 4 1
3 4 1
1 4 2
2Output
3Input
5 6 3
3 1 1
3 2 1
3 4 1
3 5 1
1 2 6
4 5 8
4Output
3Note
In the first sample the silos are located in cities 3 and 4 and on road (1, 3) at a distance 2 from city 1 (correspondingly, at a distance 1 from city 3).
In the second sample one missile silo is located right in the middle of the road (1, 2). Two more silos are on the road (4, 5) at a distance 3 from city 4 in the direction to city 5 and at a distance 3 from city 5 to city 4.
タイトル:
図を1枚与えて、図の上のすべてのs点までの距離がdのいくつかの点があることを聞きます.
一度最短絡し,sからすべての点までの最短距離を得た.次に、各エッジを列挙し、エッジに要求を満たす点があるかどうかを統計します.
1 #include <iostream>
2 #include <sstream>
3 #include <ios>
4 #include <iomanip>
5 #include <functional>
6 #include <algorithm>
7 #include <vector>
8 #include <string>
9 #include <list>
10 #include <queue>
11 #include <deque>
12 #include <stack>
13 #include <set>
14 #include <map>
15 #include <cstdio>
16 #include <cstdlib>
17 #include <cmath>
18 #include <cstring>
19 #include <climits>
20 #include <cctype>
21 using namespace std;
22 #define XINF INT_MAX
23 #define INF 0x3FFFFFFF
24 #define MP(X,Y) make_pair(X,Y)
25 #define PB(X) push_back(X)
26 #define REP(X,N) for(int X=0;X<N;X++)
27 #define REP2(X,L,R) for(int X=L;X<=R;X++)
28 #define DEP(X,R,L) for(int X=R;X>=L;X--)
29 #define CLR(A,X) memset(A,X,sizeof(A))
30 #define IT iterator
31 typedef long long ll;
32 typedef pair<int,int> PII;
33 typedef vector<PII> VII;
34 typedef vector<int> VI;
35 #define MAXN 100100
36 vector<PII> Map[MAXN];
37
38 //
39 void init() { REP(i,MAXN) Map[i].clear(); }
40
41 // s dis
42 int dis[MAXN];
43 void dijkstra(int s)
44 {
45 REP(i,MAXN){dis[i]=i==s?0:INF;}
46 int vis[MAXN] = {0};
47 priority_queue<PII, vector<PII>, greater<PII> > q;
48 q.push(MP(0,s));
49 while(!q.empty())
50 {
51 PII p = q.top(); q.pop();
52 int x = p.second;
53 if(vis[x])continue;
54 vis[x] = 1;
55 for(vector<PII>::iterator it = Map[x].begin(); it != Map[x].end(); it++)
56 {
57 int y = it->first;
58 int d = it->second;
59 if(!vis[y] && dis[y] > dis[x] + d)
60 {
61 dis[y] = dis[x] + d;
62 q.push(MP(dis[y],y));
63 }
64 }
65 }
66 }
67
68 struct node
69 {
70 int u,v,d;
71 }edge[MAXN];
72 int main()
73 {
74 ios::sync_with_stdio(false);
75 int n,m,s;
76 while(cin>>n>>m>>s)
77 {
78 int u,v,d;
79 init();
80 for(int i=0;i<m;i++)
81 {
82 cin>>u>>v>>d;
83 u--;
84 v--;
85 Map[u].PB(MP(v,d));
86 Map[v].PB(MP(u,d));
87 edge[i].u=u;
88 edge[i].v=v;
89 edge[i].d=d;
90 }
91 int l;
92 cin>>l;
93 s--;
94 dijkstra(s);
95 int ans=0;
96 for(int i=0;i<n;i++)
97 {
98 if(dis[i]==l)ans++;
99 }
100 for(int i=0;i<m;i++)
101 {
102 u=edge[i].u;
103 v=edge[i].v;
104 d=edge[i].d;
105 if(dis[u]>dis[v])swap(u,v);
106 if(dis[v]-dis[u]==d)
107 {
108 if(l>dis[u]&&l<dis[v])ans++;
109 }
110 else
111 {
112 int x=l-dis[u];
113 if(x<=0)continue;
114 if(x>d)continue;
115 if(dis[v]>l&&x<d)
116 {
117 ans++;
118 continue;
119 }
120 if(dis[v]==l&&x<d)
121 {
122 ans++;
123 continue;
124 }
125 int y=l-dis[v];
126 if(x+y==d)
127 {
128 ans++;
129 continue;
130 }
131 if(x<d-y)ans++;
132 if(y<d-x)ans++;
133 }
134 }
135 cout<<ans<<endl;
136 }
137
138
139 return 0;
140 }コード君