POJ 1986 Distance Queries 【RMQ LCA】双向存边

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题目链接

题目大意：求节点a到节点b的最短距离

思路：LCA模版，用dis数组存节点root到根节点的距离，输出 dis[a] + dis[b] - 2*dis[ LCA(a,b) ];

WA坑点：vector要双向存边，不然会WA，虽然不知道为什么要双向存边，记住就好

AC代码：

#include <iostream>#include <stdio.h>#include <math.h>#include <vector>#include <string.h>#define N 100100using namespace std;struct node{    int x,len;};vector<node> vec[N];int depth[2*N],id[N],vs[2*N],k,dp[2*N][30];long long dis[2*N];int MIN(int a, int b){    if(depth[a] <= depth[b])        return a;    return b;}void RMQ(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++)            dp[i][j] = MIN(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);}void dfs(int root, int d, int len,int f){    id[root] = k;    dis[k] = len;    vs[k] = root;    depth[k++] = d;    for(int i = 0; i < vec[root].size(); i++)    {        if(vec[root][i].x == f)            continue;        dfs(vec[root][i].x, d+1, len+vec[root][i].len,root);        vs[k] = root;        dis[k] = len;        depth[k++] = d;    }}int query(int l, int r){    if(l > r)        swap(l,r);    int k = (int)(log10(r-l+1)/log10(2));    return MIN(dp[l][k], dp[r-(1<<k)+1][k]);}int main(){    int n,m,a,b,l,in[N],q;    char ch;    while(cin>>n>>m)    {        for(int i = 0; i < N; i++)            vec[i].clear();        memset(in,0,sizeof(in));        memset(dis,0,sizeof(dis));        while(m--)        {            scanf("%d%d%d %c",&a, &b, &l, &ch);            node x = {b,l}, y = {a,l};            vec[a].push_back(x);            vec[b].push_back(y);            in[b] = 1;        }        int i = 1;        while(in[i])            i++;        k = 1;        dfs(i,0,0,-1);        RMQ(2*n-1);        cin>>q;        while(q--)        {            scanf("%d%d",&a, &b);            int fi = query(id[a],id[b]);            printf("%lld\n",dis[id[a]] - 2*dis[fi] + dis[id[b]]);        }    }    return 0;}

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