ZOJ 3195Design the city LCA

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题目：http://www.icpc.moe/onlinejudge/showProblem.do?problemCode=3195

题意：给定一个树，求三点之间的距离

思路：假定三点为u, v, w，那么(dist[u,v] + dist[v,w] + dist[u,w]) / 2就是答案

总结：wa时不要灰心，也许你离ac不远了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 50010;struct edge1{    int to, cost, next;} g1[N*2];struct edge2{    int to, ind, next;} g2[N*10];int head1[N], head2[N], dist[N], res[N*10], par[N];int cnt1, cnt2;bool vis[N];int n, m;void init(){    for(int i = 0; i < n; i++)        par[i] = i;    memset(head1, -1, sizeof head1);    memset(head2, -1, sizeof head2);    memset(vis, 0, sizeof vis);    cnt1 = cnt2 = 0;}void add_edge1(int v, int u, int c){    g1[cnt1].to = u;    g1[cnt1].cost = c;    g1[cnt1].next = head1[v];    head1[v] = cnt1++;}void add_edge2(int v, int u, int ind){    g2[cnt2].to = u;    g2[cnt2].ind = ind;    g2[cnt2].next = head2[v];    head2[v] = cnt2++;}int ser(int v){    int r = v, i = v, j;    while(r != par[r]) r = par[r];    while(i != r) j = par[i], par[i] = r, i = j;    return r;}void tarjan_lca(int v){    vis[v] = true;    int u;    for(int i = head1[v]; i != -1; i = g1[i].next)        if(!vis[u=g1[i].to])        {            dist[u] = dist[v] + g1[i].cost;            tarjan_lca(u);            par[u] = v;        }    for(int i = head2[v]; i != -1; i = g2[i].next)        if(vis[u=g2[i].to])            res[g2[i].ind] = dist[v] + dist[u] - 2 * dist[ser(u)];}int main(){    int a, b, c, x = 0;    while(~ scanf("%d", &n))    {        init();        for(int i = 0; i < n - 1; i++)        {            scanf("%d%d%d", &a, &b, &c);            add_edge1(a, b, c);            add_edge1(b, a, c);        }        scanf("%d", &m);        for(int i = 0; i < m; i++)        {            scanf("%d%d%d", &a, &b, &c);            add_edge2(a, b, i);            add_edge2(b, a, i);            add_edge2(b, c, i + m);            add_edge2(c, b, i + m);            add_edge2(a, c, i + 2 * m);            add_edge2(c, a, i + 2 * m);        }        dist[0] = 0;        tarjan_lca(0);        if(x++) printf("\n");        for(int i = 0; i < m; i++)            printf("%d\n", (res[i] + res[i+m] + res[i+2*m]) / 2);    }    return 0;}

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