cdoj 92 Journey lca

来源：互联网发布：双色球出号绝密算法编辑：程序博客网时间：2024/06/14 11:50

题意：给定一棵树，然后加一条边，有若干询问，问你每一个询问（u,v），加了这条边后可以从u到v节省多少距离。

思路：一共三种情况， 1，原路

2，u - x - y - v

3，u - y - x - v

#include <bits/stdc++.h>using namespace std;const int MAXN = 100010;struct node{int to, w, next;}edge[MAXN*2];int tot,head[MAXN];void init(){tot = 0;memset(head, -1, sizeof(head));}void add_edge(int u, int v, int w){edge[tot].to = v;edge[tot].w = w;edge[tot].next = head[u];head[u] = tot++;}//LCA部分int rmq[2*MAXN];//rmq数组，就是欧拉序列对应的深度序列struct ST{    int mm[2*MAXN];    int dp[2*MAXN][20];//最小值对应的下标    void init(int n)    {        mm[0] = -1;        for(int i = 1;i <= n;i++)        {            mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];            dp[i][0] = i;        }        for(int j = 1; j <= mm[n];j++)            for(int i = 1; i + (1<<j) - 1 <= n; i++)                dp[i][j] = rmq[dp[i][j-1]] < rmq[dp[i+(1<<(j-1))][j-1]]?dp[i][j-1]:dp[i+(1<<(j-1))][j-1];    }    int query(int a,int b)//查询[a,b]之间最小值的下标    {        if(a > b)swap(a,b);        int k = mm[b-a+1];        return rmq[dp[a][k]] <= rmq[dp[b-(1<<k)+1][k]]?dp[a][k]:dp[b-(1<<k)+1][k];    }};int F[MAXN*2];//欧拉序列，就是dfs遍历的顺序，长度为2*n-1,下标从1开始int P[MAXN];//P[i]表示点i在F中第一次出现的位置int cnt;int dis[MAXN]; ST st;void dfs(int u,int pre,int dep, int d){dis[u] = d;    F[++cnt] = u;    rmq[cnt] = dep;    P[u] = cnt;    for(int i = head[u];i != -1;i = edge[i].next)    {        int v = edge[i].to;        if(v == pre)continue;        dfs(v,u,dep+1, edge[i].w+d);        F[++cnt] = u;        rmq[cnt] = dep;    }}void LCA_init(int root,int node_num)//查询LCA前的初始化{    cnt = 0;    dfs(root,root,0, 0);    st.init(2*node_num-1);}int query_lca(int u,int v)//查询u,v的lca编号{    return F[st.query(P[u],P[v])];}int calc(int u, int v){int LCA = query_lca(u, v);return dis[u] + dis[v] - 2*dis[LCA];}int main(){int T;cin>>T;int icase = 0;while(T--){init();printf("Case #%d:\n", ++icase);int n,q;scanf("%d %d", &n, &q);for(int i=1; i<n; i++){int u,v,w;scanf("%d %d %d", &u, &v, &w);add_edge(u, v, w), add_edge(v, u, w);}LCA_init(1, n);int u,v,w;int lca = query_lca(u, v);scanf("%d %d %d", &u, &v, &w);while(q--){int x,y;scanf("%d %d", &x, &y);int sum1 = calc(x, y);int sum2 = calc(x, u) + w + calc(y, v);int sum3 = calc(y, u) + w + calc(x, v);if(sum2 >= sum1 && sum3>=sum1) printf("0\n");else printf("%d\n", sum1 - min(sum2, sum3));}}return 0;}

0 0