2015沈阳网络赛 C

树上差分 用到了树链剖分的方法求 lca。

这题数据有点弱呀，

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 20020, M = 2 * 200020;

int n, m;
int h[N], e[M], ne[M], idx;
int dfs[N], cha[N], d[N], sz[N], fa[N], son[N], bl[N], cnt;
int ans;

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

void dfs1(int x) {
    sz[x] = 1;
    d[x] = d[fa[x]] + 1;
    for(int i = h[x]; ~i; i = ne[i]) {
        int j = e[i];
        if(j != fa[x]){
            fa[j] = x;
            dfs1(j);
            sz[x] += sz[j];
            if(sz[son[x]] < sz[j]) son[x] = j;
        }
    }
}

void dfs2(int x, int belong) {
    dfs[x] = ++cnt;
    bl[x] = belong;
    if(son[x]) dfs2(son[x], belong);
    for(int i = h[x]; ~i; i = ne[i]) {
        int j = e[i];
        if(j != son[x] && j != fa[x]) {
            dfs2(j, j);
        }
    }
}

int lca(int l, int r) {
    while(bl[l] != bl[r]) {
        if(d[bl[l]] > d[bl[r]]) swap(l, r);
        r = fa[bl[r]];
    }
    return d[l] > d[r] ? r : l;
}

void dfs_ans(int x) {
    for(int i = h[x];~i;i = ne[i]) {
        int j = e[i];
        if(j != fa[x]){
            dfs_ans(j);
            cha[x] += cha[j];
        }
    }
    if(x != 1)//边权写法
        ans = min(ans, cha[x]);
}

int main(){
    int _, l, r;
    scanf("%d",&_);
    for(int cas = 1; cas <= _;cas ++) {
        idx = 0;
        cnt = 0;
        ans = 0x7f7f7f7f;
        memset(h, -1, sizeof h);
        memset(son, 0, sizeof son);
        memset(cha ,0, sizeof cha);

        scanf("%d%d",&n,&m);
        for(int i = 0;i < n-1;i ++) {
            scanf("%d%d",&l, &r) ;
            add(l, r), add(r, l);
        }
        dfs1(1);
        dfs2(1, 1);
        for(int i = 0;i < m - n + 1;i ++) {
            scanf("%d%d",&l,&r);
            int f = lca(l, r);
            // cha[l] ++, cha[r] ++, cha[f] --, cha[fa[f]] --;//点权写法
            cha[l] ++, cha[r] ++, cha[f] -= 2;//边权写法
        }
        dfs_ans(1);
        printf("Case #%d: %d\n", cas, ans + 1);
    }
    return 0;
}
/*
1
4 5
1 2
2 3
3 4
1 3
1 4
*/