383. 观光

思路

仿照第二短路的解法而已,额外添加了计数功能而已,代码很好理解。

AC code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

struct node{
int ver, type, dist;
bool operator > (const node& nod) const{
return dist > nod.dist;
}
};

const int N = 1010, M = 10010;
int h[N], e[M], ne[M], w[M], idx;
int dist[N][2], cnt[N][2];
bool st[N][2];
int t, n, m, S, T;

void add(int a, int b, int c){
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int dijkstra(){
memset(st, false, sizeof(st));
memset(dist, 0x3f, sizeof(dist));
memset(cnt, 0, sizeof(cnt));

dist[S][0] = 0, cnt[S][0] = 1;
priority_queue<node, vector<node>, greater<node>> q;
q.push({S, 0, 0});

while(q.size()){
node tt = q.top();
q.pop();
int ver = tt.ver, type = tt.type, dis = tt.dist;
int count = cnt[ver][type];
if(st[ver][type]) continue;
st[ver][type] = true;
for(int i = h[ver]; ~i; i = ne[i]){
int j = e[i];
if(dist[j][0] > dis + w[i]){
dist[j][1] = dist[j][0], cnt[j][1] = cnt[j][0];
q.push({j, 1, dist[j][1]});
dist[j][0] = dis + w[i], cnt[j][0] = count;
q.push({j, 0, dist[j][0]});
}
else if(dist[j][0] == dis + w[i]) cnt[j][0] += count;
else if(dist[j][1] > dis + w[i]){
dist[j][1] = dis + w[i], cnt[j][1] = count;
q.push({j, 1, dist[j][1]});
}
else if(dist[j][1] == dis + w[i]) cnt[j][1] += count;
}
}
int res = cnt[T][0];
if(dist[T][0] + 1 == dist[T][1]) res += cnt[T][1];
return res;
}

int main(){
cin >> t;
while(t--){
cin >> n >> m;
memset(h, -1, sizeof(h));
idx = 0;
for(int i = 0; i < m; i++){
int a, b, c;
cin >> a >> b >> c;
add(a, b, c);
}
cin >> S >> T;
cout << dijkstra() << endl;
}
return 0;
}