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 + 3;
const int maxn = 1e6 + 5;
const int maxm = 2e6 + 5;
int head[maxn];
struct edge{
    int to, nxt;
}e[maxm * 2];
int tot = 0;

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[1] = true;
    cnt[1] = 1;
    d[1] = 0;
    que.push(make_pair(d[1], 1));
    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] + 1){
                d[e[i].to] = d[now.second] + 1;
                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] + 1){
                cnt[e[i].to] = (cnt[e[i].to] + cnt[now.second]) % mod;
            }
        }
    }
    
}
 
int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 0; i < m; i++){
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y);
    }
    dijkstra();
    for(int i = 1; i <= n; i++){
        printf("%d
", cnt[i]);
    }
}