洛谷P1144 最短路计数【堆优化dijkstra】

题目：https://www.luogu.org/problemnew/show/P1144

题意：问1到各个节点的最短路有多少条。

思路：如果松弛的时候发现是相等的，说明可以经过该点的最短路径到达当前点，也就是说最短路径变多了。

 #include<cstdio>
 #include<cstdlib>
 #include<map>
 #include<set>
 #include<cstring>
 #include<algorithm>
 #include<vector>
 #include<cmath>
 #include<stack>
 #include<queue>
 #include<iostream>

 #define inf 0x3f3f3f3f
 using namespace std;
 typedef long long LL;
 typedef pair<int, int> pr;

 int n, m;
 const int mod = 1e5 + ;
 const int maxn = 1e6 + ;
 const int maxm = 2e6 + ;
 int head[maxn];
 struct edge{
     int to, nxt;
 }e[maxm * ];
 int tot = ;

 void add(int x, int y)
 {
     e[++tot].to = y;
     e[tot].nxt = head[x];
     head[x] = tot;
     e[++tot].to = x;
     e[tot].nxt = head[y];
     head[y] = tot;
 }

 struct cmp{
     bool operator()(const pr &a, const pr &b)
     {
         return a.first > b.first;
     }
 }; 

 int d[maxn], cnt[maxn];
 bool vis[maxn];
 void dijkstra()
 {
     memset(d, 0x3f, sizeof(d));
     priority_queue<pr>que;
     vis[] = true;
     cnt[] = ;
     d[] = ;
     que.push(make_pair(d[], ));
     while(!que.empty()){
         pr now = que.top();que.pop();
         vis[now.second] = true;
         for(int i = head[now.second]; i; i = e[i].nxt){
             if(d[e[i].to] > d[now.second] + ){
                 d[e[i].to] = d[now.second] + ;
                 cnt[e[i].to] = cnt[now.second];
                 que.push(make_pair(-d[e[i].to], e[i].to));
             }
             else if(d[e[i].to] == d[now.second] + ){
                 cnt[e[i].to] = (cnt[e[i].to] + cnt[now.second]) % mod;
             }
         }
     }

 }

 int main()
 {
     scanf("%d%d", &n, &m);
     for(int i = ; i < m; i++){
         int x, y;
         scanf("%d%d", &x, &y);
         add(x, y);
     }
     dijkstra();
     for(int i = ; i <= n; i++){
         printf("%d\n", cnt[i]);
     }
 }