南京DAY ？

A,B,C,D跳过

E比较复杂的宽搜，没return 喜提五分,100–>5

F还行的DP，因为E题return 改不出来导致比赛结束前交，没评测到，100–>0

aaaaaaaaaaaa

G题理解了就很好理解了，没理解还要好好理解一下（？

（因为不知名原因80分，TLE 2014/2000ms）

代码：

#include <bits/stdc++.h>
using namespace std;
int n, q, aa, bb, f[1000005], dp[1000005],  MLE, RE, color[1000005],lastans;
struct sd {
    int num, kl;
} sum[1000005][4];
vector<int> G[1000005];
int cmp(sd A, sd B) { return A.num < B.num; }
void dfs(int x, int f, int cc) {
    color[x] = cc;
    int top = 0;
    int hu=0;
    for (int i = 0; i < G[x].size(); i++) {
        if (G[x][i] != f) {
            if (x == 1) {
                cc = G[x][i];
            }
            dfs(G[x][i], x, cc);
            if(top<=2) 
            {
                top++;
                sum[x][top]={min(sum[G[x][i]][1].num, G[x][i]), G[x][i] };
            }
            else
            {
                hu=min(sum[G[x][i]][1].num, G[x][i]);
                sort(sum[x]+1,sum[x]+3+1,cmp);
                if(hu<sum[x][3].num)
                {
                    sum[x][3].num=hu;
                    sum[x][3].kl=G[x][i];
                    sort(sum[x]+1,sum[x]+3+1,cmp);
                }

            }
            //  if(x==2) cout<<ls[top-1].num<<endl;
        }
    }

    return;
}
int qwq(int hhh) {
    if (hhh == 0)
        return 11451419;
    else
        return hhh;
}
void dfs2(int x, int f, int last) {
    if (x == 1) {
        for (int i = 0; i < G[x].size(); i++) {
            if (G[x][i] != f) {
                dfs2(G[x][i], x, last);
            }
        }
    } else {
        int TLE;
        dp[x] = min(last, qwq(sum[x][1].num));
        for (int i = 0; i < G[x].size(); i++) {
            TLE = last;
            if (G[x][i] != f) {
                if (sum[x][1].kl != G[x][i]) {
                    TLE = min(last, qwq(sum[x][1].num));
                } else {
                    if (sum[x][2].kl != G[x][i]) {
                        TLE = min(last, qwq(sum[x][2].num));
                    } else {
                        if (sum[x][3].kl != G[x][i]) {
                            TLE = min(last, qwq(sum[x][3].num));
                        }
                    }
                }
                dfs2(G[x][i], x, TLE);
            }
        }
    }
}
int main() {
    while(~scanf("%d%d",&n,&q))
    {
    lastans=0;

    for (int i = 1; i <= n; i++) {
        dp[i] = sum[i][1].num = sum[i][2].num = sum[i][3].num = 11451419;
        G[i].clear();
    }
    for (int i = 1; i <= n - 1; i++) {
        scanf("%d%d",&aa,&bb);
        G[aa].push_back(bb);
        G[bb].push_back(aa);
    }
    dfs(1, 0, 0);
    dfs2(1, 0, 11451419);
    for (int i = 1; i <= q; i++) {
        scanf("%d%d",&aa,&bb);
        aa=aa^lastans,bb=bb^lastans;
        if (color[aa] == color[bb])
            {
                printf("%d\n",1);
                lastans=1;
            }
        else {
            MLE = 11451419;
            if (sum[1][1].kl != color[aa] && sum[1][1].kl != color[bb]) {
                MLE = qwq(sum[1][1].num);
            } else {
                if (sum[1][2].kl != color[aa] && sum[1][2].kl != color[bb]) {
                    MLE = qwq(sum[1][2].num);
                } else {
                    if (sum[1][3].kl != color[aa] && sum[1][3].kl != color[bb]) {
                        MLE = qwq(sum[1][3].num);
                    }
                }
            }
            MLE = min(MLE, min(dp[aa], dp[bb]));
            if (MLE == 11451419)
                {
                    printf("%d\n",n); ;
                    lastans=n;
                }
            else
            {
                printf("%d\n",MLE);
                lastans=MLE;
            }

        }
    }
        }

    return 0;
}