AcWing 349 Dijkstra

Title

Portal AcWing 349 Dark Castle

answer

Count the number of shortest path spanning trees in the graph. Consider the process of spanning tree, take node $1$ is the root, keep adding the node with the shortest shortest path among the nodes that have not joined the spanning tree$i$ , this time suretyiiThe predecessor nodes of the shortest path of$i$ must all be in the spanning tree.

Use $DIJKSTR\; \&\; ltA$ Solution single source shortest, statistics of the number of nodes in the shortest predecessor nodes, i.e. multiplied the answer in accordance with the principle of multiplication. The picture is a dense graph, using the plain$Dijkstra$ , total time complexity$O(N^2)$。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1005;
const ll mod = (1LL << 31) - 1;
int N, M, G[maxn][maxn], ds[maxn], cnt[maxn];
bool vs[maxn];

void Dijkstra(int s)
{
    
    
    memset(ds, 0x3f, sizeof(ds));
    memset(vs, 0, sizeof(vs));
    ds[s] = 0;
    for (int i = 1; i < N; ++i)
    {
    
    
        int x = 0;
        for (int j = 1; j <= N; ++j)
            if (!vs[j] && (!x || ds[j] < ds[x]))
                x = j;
        vs[x] = 1;
        for (int j = 1; j <= N; ++j)
            ds[j] = min(ds[j], ds[x] + G[x][j]);
    }
}

int main()
{
    
    
    scanf("%d%d", &N, &M);
    memset(G, 0x3f, sizeof(G));
    for (int i = 1; i <= N; ++i)
        G[i][i] = 0;
    for (int i = 1, x, y, z; i <= M; ++i)
        scanf("%d%d%d", &x, &y, &z), G[x][y] = G[y][x] = min(G[x][y], z);
    Dijkstra(1);
    ll res = 1;
    for (int i = 1; i <= N; ++i)
        for (int j = 1; j <= N; ++j)
            if (i != j && ds[i] + G[i][j] == ds[j])
                ++cnt[j];
    for (int i = 2; i <= N; ++i)
        res = (res * cnt[i]) % mod;
    printf("%lld\n", res);
    return 0;
}