传送门：https://www.luogu.org/problemnew/show/P1144
这虽然是一道普及+的题，然而我发现我现在还没做过，这也就直接导致我今天模拟T2只杠了个暴力分……
那这道题怎么做呢？既然是最短路，那么一定要用spfa或dijkstra了，这里就讲dijkstra的做法吧，主要是这比spfa简单点，而且spfa不是那啥了吗。
众所周知，在那个所谓的什么松弛操作的时候，如果dis[u] + c[u->v] < dis[v]，我们就更新dis[v]，这时候从u过来的路径，就可能是到v的最短路，所以到v的最短路条数num[v] = num[u];而另一种情况是dis[u] + c[u->v] == dis[v]时，那么num[v] += num[u]了。嗯，完事了
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter printf(" ") 13 #define space printf(" ") 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const int eps = 1e-8; 19 const int maxn = 1e6 + 5; 20 const int mod = 100003; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) 27 { 28 ans = ans * 10 + ch - '0'; ch = getchar(); 29 } 30 if(last == '-') ans = -ans; 31 return ans; 32 } 33 inline void write(ll x) 34 { 35 if(x < 0) x = -x, putchar('-'); 36 if(x >= 10) write(x / 10); 37 putchar(x % 10 + '0'); 38 } 39 40 int n, m; 41 vector<int> v[maxn]; 42 43 #define pr pair<int, int> 44 #define mp make_pair 45 priority_queue<pr, vector<pr>, greater<pr> > q; 46 int dis[maxn], num[maxn]; 47 bool done[maxn]; 48 void dijkstra(int s) 49 { 50 for(int i = 1; i <= n; ++i) dis[i] = INF; 51 dis[s] = 0; num[s] = 1; 52 q.push(mp(dis[0], s)); 53 while(!q.empty()) 54 { 55 int now = q.top().second; q.pop(); 56 if(done[now]) continue; 57 done[now] = 1; 58 for(int i = 0; i < (int)v[now].size(); ++i) 59 { 60 if(dis[now] + 1 < dis[v[now][i]]) //此题边权都是1 61 { 62 dis[v[now][i]] = dis[now] + 1; 63 num[v[now][i]] = num[now]; 64 q.push(mp(dis[v[now][i]], v[now][i])); 65 } 66 else if(dis[now] + 1 == dis[v[now][i]]) num[v[now][i]] = (num[v[now][i]] + num[now]) % mod; 67 } 68 } 69 } 70 71 int main() 72 { 73 n = read(); m = read(); 74 for(int i = 1; i <= m; ++i) 75 { 76 int x = read(), y = read(); 77 v[x].push_back(y); v[y].push_back(x); 78 } 79 dijkstra(1); 80 for(int i = 1; i <= n; ++i) {write(num[i]); enter;} 81 return 0; 82 }