ACM-ICPC 2018 徐州赛区网络预赛 J Maze Designer（最大生成树+LCA）

https://nanti.jisuanke.com/t/31462

题意

一个N*M的矩形,每个格点到其邻近点的边有其权值,需要构建出一个迷宫,使得构建迷宫的边权之和最小,之后Q次查询,每次给出两点坐标,给出两点之间的最短路径

分析

可以把每个格点视作视作图的点,隔开两点的边视作图的边,则构建迷宫可以视作求其生成树,剩余的边就是组成迷宫的墙.因为要花费最小,所以使删去的墙权置最大即可,呢么就是求最大生成树即可.
然后每次查询相当于查这个最大生成树上任意两点的最短距离,到这一步就是个LCA了。

这题码量不少，还好队友给力。

#include <iostream>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <map>
#include <algorithm>
#include <queue>
#include <set>
#include <cmath>
#include <sstream>
#include <stack>
#include <fstream>
#include <ctime>
#pragma warning(disable:4996);
#define mem(sx,sy) memset(sx,sy,sizeof(sx))
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-8;
const double PI = acos(-1.0);
const ll llINF = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
using namespace std;
//#define pa pair<int, int>
//const int mod = 1e9 + 7;
const int maxn = 1000005;
const int maxq = 300005;
struct node {
    int u, v, w, next, lca;
};
 
struct LCA {
    node edges[maxn], ask[maxq];
    int ghead[maxn], gcnt, ahead[maxn], acnt;
    int anc[maxn];
    int vis[maxn];
    ll dist[maxn];
    int fa[maxn];
 
    void addedge(int u, int v, int w) {
        edges[gcnt].v = v;
        edges[gcnt].w = w;
        edges[gcnt].next = ghead[u];
        ghead[u] = gcnt++;
    }
    
    void addask(int u, int v) {
        ask[acnt].u = u;
        ask[acnt].v = v;
        ask[acnt].next = ahead[u];
        ahead[u] = acnt++;
    }
    
    void init() {
        mem(vis, 0);
        mem(ghead, -1);
        mem(ahead, -1);
        gcnt = 0;
        acnt = 0;
    }
 
    int Find(int x) {
        return fa[x] == x ? x : (fa[x] = Find(fa[x]));
    }
 
    void getLCA(int u, int d) {
        dist[u] = d;
        fa[u] = u;
        vis[u] = 1;
        for (int i = ghead[u]; i != -1; i = edges[i].next) {
            int v = edges[i].v;
            if (!vis[v]) {
                getLCA(v, d + edges[i].w);
                fa[v] = u;
                anc[fa[v]] = u;
            }
        }
        for (int i = ahead[u]; i != -1; i = ask[i].next) {
            int v = ask[i].v;
            if (vis[v])
                ask[i].lca = ask[i ^ 1].lca = Find(ask[i].v);
        }
 
    }
 
}L;
 
struct edge {
    int u, v;
    ll w;
    bool operator<(const edge &e)const { return w>e.w; }
    edge(int _u = 0, int  _v = 0, ll _w = 0)
        :u(_u), v(_v), w(_w) {}
};
 
struct Kruskal {
    int n, m;
    edge edges[maxn];
    int fa[maxn];
    int Find(int x) {
        return fa[x] == -1 ? x : fa[x] = Find(fa[x]);
    }
    void init(int _n) {
        this->n = _n;
        m = 0;
        mem(fa, -1);
    }
 
    void AddEdge(int u, int v, ll dist) {
        edges[m++] = edge(u, v, dist);
    }
 
    ll kruskal() {
        ll sum = 0;
        int cntnum = 0;
        sort(edges, edges + m);
        for (int i = 0; i < m; i++) {
            int u = edges[i].u, v = edges[i].v;
            if (Find(u) != Find(v)) {
                L.addedge(u, v, 1);
                L.addedge(v, u, 1);
                //cout << u << " " << v << endl;
                sum += edges[i].w;
                fa[Find(u)] = Find(v);
                if (++cntnum >= n - 1) return sum;
            }
        }
        return -1;
    }
}G;
 
int main() {
    int n, m;
    while (~scanf("%d%d", &n, &m)) {
        G.init(n*m);
        L.init();
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                int w1, w2; char c1, c2;
                scanf(" %c%d %c%d", &c1, &w1, &c2, &w2);
                if (c1 == 'D') {
                    G.AddEdge((i - 1)*m + j, i*m + j, w1);
                }
                if (c2 == 'R') {
                    G.AddEdge((i - 1)*m + j, (i - 1)*m + j + 1, w2);
                }
            }
        }
        G.kruskal();
        int q;
        scanf("%d", &q);
        for (int i = 1, x1, x2, y1, y2; i <= q; i++) {
            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
            int u = (x1 - 1)*m + y1;
            int v = (x2 - 1)*m + y2;
            L.addask(u, v);
            L.addask(v, u);
        }
        L.getLCA(1, 0);
        for (int i = 0; i < L.acnt; i += 2) {
            printf("%lld
", L.dist[L.ask[i].u] + L.dist[L.ask[i].v] - 2 * L.dist[L.ask[i].lca]);
        }
    }
}