标签：tmp   between   for   tin   can   枚举   dir   esc   解决   

题目

You are given a weighed undirected connected graph, consisting of n vertices and m edges.

You should answer q queries, the i-th query is to find the shortest distance between vertices \(u＿i\) and \(v_i\).

输入格式

The first line contains two integers \(n\) and \(m (1≤n,m≤105,m?n≤20)\) — the number of vertices and edges in the graph.

Next m lines contain the edges: the i-th edge is a triple of integers \(v_i,u_i,d_i (1≤u_i,v_i≤n,1≤d_i≤10^9,u_i≠v_i)\). This triple means that there is an edge between vertices ui and vi of weight di. It is guaranteed that graph contains no self-loops and multiple edges.

The next line contains a single integer \(q (1≤q≤10^5)\) — the number of queries.

Each of the next q lines contains two integers \(u_i\) and \(v_i (1≤u_i,v_i≤n)\) — descriptions of the queries.

Pay attention to the restriction \(m?n ≤ 20\).

输出格式

Print q lines.

The i-th line should contain the answer to the i-th query — the shortest distance between vertices \(u_i\) and \(v_i\).

输入样例1

3 3
1 2 3
2 3 1
3 1 5
3
1 2
1 3
2 3

输出样例1

3
4
1

输入样例2

8 13
1 2 4
2 3 6
3 4 1
4 5 12
5 6 3
6 7 8
7 8 7
1 4 1
1 8 3
2 6 9
2 7 1
4 6 3
6 8 2
8
1 5
1 7
2 3
2 8
3 7
3 4
6 8
7 8

输出样例2

7
5
6
7
7
1
2
7

题解

由于\(m?n ≤ 20\), 边和点的数量接近, 所以大部分最短路用lca解决, 剩下的一些没有计算的边, 对它们的顶点重新跑一遍最短路, 然后枚举每一个点, 更新两个目标点之间的最短距离.

代码

#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const long long INF = 1e18;
const int N = 100005;
int head[N], tote, f[N][21], deth[N], tmp[100], tot, n, m;
long long dis[N], d[50][N];
bool vis[N];
struct Edge { int to, next, value; } edges[N << 1];
void add(int u, int v, int w) {
    edges[++tote]= (Edge){ v, head[u], w}, head[u] = tote;
    edges[++tote]= (Edge){ u, head[v], w}, head[v] = tote;
}
priority_queue<pair<long long, int>, vector<pair<long long, int> >, greater<pair<long long, int> > > q;
void dijkstra(int id, int S) {
    for (int i = 1; i <= n; ++i) dis[i] = INF, vis[i] = false;
    dis[S] = 0;
    q.push(make_pair(0, S));
    while (!q.empty()) {
        int u = q.top().second;
        q.pop();
        if (vis[u]) continue;
        vis[u] = true;
        for (int i = head[u]; i; i = edges[i].next) {
            int v = edges[i].to;
            if (dis[v] > dis[u] + edges[i].value) {
                dis[v] = dis[u] + edges[i].value;
                q.push(make_pair(dis[v], v));
            }
        }
    }
    for (int i = 1; i <= n; ++i) d[id][i] = dis[i];
}
void dfs(int u, int fa) {
    vis[u] = true;
    f[u][0] = fa;
    deth[u] = deth[fa] + 1;
    for (int i = head[u]; i; i = edges[i].next) {
        int v = edges[i].to;
        if (v == fa) continue;
        if (vis[v]) tmp[++tot] = u, tmp[++tot] = v;
        else dis[v] = dis[u] + edges[i].value, dfs(v, u);
    }
}
int lca(int u, int v) {
    if (deth[u] < deth[v]) swap(u, v);
    int d = deth[u] - deth[v];
    for (int i = 20; i >= 0; --i)
        if (d & (1 << i)) u = f[u][i];
    if (u == v) return u;
    for (int i = 20; i >= 0; --i)
        if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];
    return f[u][0];
}
inline int input() { int t; scanf("%d", &t); return t; }
int main() {
    n = input(), m = input();
    for (int i = 1; i <= m; ++i){
        int u = input(), v = input(), w = input();
        add(u, v, w);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; ++i) d[0][i] = dis[i];
    for (int j = 1; j <= 20; ++j)
        for (int i = 1; i <= n; ++i) f[i][j] = f[f[i][j - 1]][j - 1];
    sort(tmp + 1, tmp + tot + 1);
    int j = 1;
    for (int i = 2; i <= tot; ++i)
        if (tmp[j] != tmp[i]) tmp[++j] = tmp[i];
    for (int i = 1; i <= j; ++i) dijkstra(i, tmp[i]);
    for(int q = input(); q; q--) {
        int u = input(), v = input();
        long long ans = d[0][u] + d[0][v] - 2 * d[0][lca(u, v)];
        for (int i = 1; i <= j; ++i) ans = min(ans, d[i][u] + d[i][v]);
        printf("%lld\n", ans);
    }
    return 0;
}

CF1051F The Shortest Statement

标签：tmp   between   for   tin   can   枚举   dir   esc   解决   

原文地址：https://www.cnblogs.com/youxam/p/cf1051F.html