hdu2874(lca / tarjan离线 + RMQ在线)

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2874

题意: 给出 n 个顶点 m 条边的一个森林, 有 k 个形如 x y 的询问, 输出 x, y 之间的最短路径.

思路: 如果将森林换成一棵树的话就是一道 lca 模板题了, 不过本题需要稍作改动.

解法1: tarjan

只需要先判断一下 x, y 是否在一颗树里面就 ok 了, 不过这道题的询问有点多, 很容易 mle.

代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

const int MAXN = 1e4 + 10;
const int MAX = 1e6 + 10;
struct node1{
    int v, w, next;
}edge1[MAXN << 1];

struct node2{
    int u, v, next;
}edge2[MAX << 1];//edge1记录树, edge2记录询问

int vis[MAXN], pre[MAXN], dis[MAXN], sol[MAX];//vis[i]标记i是否已经搜索过, pre[i]记录i的根节点, dis[i]记录i到根节点的距离
int head1[MAXN], head2[MAXN], ip1, ip2, cnt;

void init(void){
    memset(sol, 0, sizeof(sol));
    memset(vis, 0, sizeof(vis));
    memset(dis, 0, sizeof(dis));
    memset(head1, -1, sizeof(head1));
    memset(head2, -1, sizeof(head2));
    ip1 = ip2 = cnt = 0;
}

void addedge1(int u, int v, int w){//前向星
    edge1[ip1].v = v;
    edge1[ip1].w = w;
    edge1[ip1].next = head1[u];
    head1[u] = ip1++;
}

void addedge2(int u, int v){
    edge2[ip2].u = u;
    edge2[ip2].v = v;
    edge2[ip2].next = head2[u];
    head2[u] = ip2++;
}

int find(int x){
    return pre[x] == x ? x : pre[x] = find(pre[x]);
}

void jion(int x, int y){
    x = find(x);
    y = find(y);
    if(x != y) pre[y] = x;
}

void tarjan(int u){
    pre[u] = u;
    vis[u] = cnt;//将不同的联通块标记成不同的数字
    for(int i = head1[u]; i != -1; i = edge1[i].next){
        int v = edge1[i].v;
        int w = edge1[i].w;
        if(!vis[v]){
            dis[v] = dis[u] + w;
            tarjan(v);
            jion(u, v);
        }
    }
    for(int i = head2[u]; i != -1; i = edge2[i].next){
        int v = edge2[i].v;
        if(vis[v] == cnt) sol[i >> 1] = find(v);//只有同一个联通块的顶点才存在lca
    }
}

int main(void){
    int t, n, m, x, y, z;
    while(~scanf("%d%d%d", &n, &m, &t)){
        init();
        for(int i = 0; i < m; i++){
            scanf("%d%d%d", &x, &y, &z);
            addedge1(x, y, z);
            addedge1(y, x, z);
        }
        for(int i = 0; i < t; i++){
            scanf("%d%d", &x, &y);
            addedge2(x, y);
            addedge2(y, x);
        }
        for(int i = 1; i <= n; i++){
            if(!vis[i]){
                cnt++;
                dis[i] = 0;//顶点到自己的距离为0
                tarjan(i);
            }
        }
        for(int i = 0; i < t; i++){
            int cc = i << 1;
            int u = edge2[cc].u;
            int v = edge2[cc].v;
            int lca = sol[i];//sol开两倍空间会超内存
            if(!lca) puts("Not connected");
            else printf("%d
", dis[u] + dis[v] - 2 * dis[lca]);
        }
    }
    return 0;
}

解法2: lca 转 RMQ

可以先拟个虚根, 另外输出前先判一下 x, y 是否在同一棵树里面即可.

代码:

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>
using namespace std;

const int inf = 1e9;
const int MAXN = 4e4 + 10;
struct node{
    int v, w, next;
}edge[MAXN << 1];

int dp[MAXN << 1][30]; //dp[i][j]存储deep数组中下标i开始长度为2^j的子串中最小值的下标
int first[MAXN], ver[MAXN << 1], deep[MAXN << 1];
int vis[MAXN], head[MAXN], dis[MAXN], tag[MAXN], ip, indx, cnt;

inline void init(void){
    memset(tag, 0, sizeof(tag));
    memset(vis, 0, sizeof(vis));
    memset(head, -1, sizeof(head));
    ip = 0;
    indx = 0;
    cnt = 0;
}

void addedge(int u, int v, int w){
    edge[ip].v = v;
    edge[ip].w = w;
    edge[ip].next = head[u];
    head[u] = ip++;
}

void dfs(int u, int h){
    vis[u] = 1; //标记已搜索过的点
    ver[++indx] = u; //记录dfs路径
    first[u] = indx; //记录顶点u第一次出现时对应的ver数组的下标
    deep[indx] = h; //记录ver数组中对应下标的点的深度
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].v;
        if(!vis[v]){
            dis[v] = dis[u] + edge[i].w;
            dfs(v, h + 1);
            ver[++indx] = u;
            deep[indx] = h;
        }
    }
}

void dfs1(int u){
    tag[u] = cnt;
    for(int i = head[u]; i != -1; i = edge[i].next){
        if(!tag[edge[i].v]) dfs1(edge[i].v);
    }
}

void ST(int n){
    for(int i = 1; i <= n; i++){
        dp[i][0] = i;
    }
    for(int j = 1; (1 << j) <= n; j++){
        for(int i = 1; i + (1 << j) - 1 <= n; i++){
            int x = dp[i][j - 1], y = dp[i + (1 << (j - 1))][j - 1];
            dp[i][j] = deep[x] < deep[y] ? x : y;
        }
    }
}

int RMQ(int l, int r){
    int len = log2(r - l + 1);
    int x = dp[l][len], y = dp[r - (1 << len) + 1][len];
    return deep[x] < deep[y] ? x : y;
}

int LCA(int x, int y){
    int l = first[x], r = first[y];
    if(l > r) swap(l, r);
    int pos = RMQ(l, r);
    return ver[pos];
}

int main(void){
    int n, m, k, x, y, z;
    while(~scanf("%d%d%d", &n, &m, &k)){
        init();
        for(int i = 0; i < m; i++){
            scanf("%d%d%d", &x, &y, &z);
            addedge(x, y, z);
            addedge(y, x, z);
        }
        for(int i = 1; i <= n; i++){
            if(!tag[i]){
                cnt++;
                dfs1(i);
                addedge(i, 0, inf); //拟个虚根0
                addedge(0, i, inf);
            }
        }
        dis[0] = 0;
        dfs(0, 1);
        ST(2 * n - 1);
        for(int i = 0; i < k; i++){
            scanf("%d%d", &x, &y);
            int lca = LCA(x, y);
            if(tag[x] != tag[y]) puts("Not connected");
            else printf("%d
", dis[x] + dis[y] - 2 * dis[lca]);
        }
    }
    return 0;
}