SPOJ QTREE4 Query on a tree IV(边分治)

看这里

边分治,用堆维护两边的分治的子树(感觉这两个堆的更新用的很巧妙, 然后用链表维护每个点所在的分治结构). 顺便留一份边分治模板

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

#define LL long long 
#define pii pair
#define MP make_pair
#define inf 0x3f3f3f3f
#define N 200010
#define M 4000020

int fst[N], vv[M], nxt[M], cost[M], e;
int f[N], rt, edg, C[N];
priority_queue<int> q1[N], q2[N];
int sz[N], mx[N], tot, tim;
int col[N], lc[N], rc[N], ans[N];
int x[N], cx;
bool vis[M];

void init(){
    memset(fst, -1, sizeof fst);
    memset(f, -1, sizeof f);
    e = 0;
}
void _add(int u, int v, int c){
    vv[e] = v, nxt[e] = fst[u], cost[e] = c, fst[u] = e++;
}
void add(int u, int v, int c){
    vv[e] = v, nxt[e] = f[u], cost[e] = c, f[u] = e++;
}

void dfsS(int u, int p){
    sz[u] = 1;
    for(int i = fst[u]; ~i; i = nxt[i]){
        int v = vv[i];
        if(v == p || vis[i]) continue;
        dfsS(v, u);
        sz[u] += sz[v];
    }
}
void dfsE(int u, int p, int d){
    add(u, tim, d);
    x[cx++] = d;
    for(int i = fst[u]; ~i; i = nxt[i]){
        int v = vv[i], c = cost[i];
        if(v == p || vis[i]) continue;
        mx[v] = max(sz[v], tot - sz[v]);
        if(rt == -1 || mx[v] < mx[rt])
            rt = v, edg = i, C[tim] = c;
        dfsE(v, u, d + c);
    }
}   
pii find_edge(int u){
    dfsS(u, -1);
    tot = sz[u], rt = -1, edg = -1, cx = 0;
    dfsE(u, -1, 0); 
    for(int i = 0; i < cx; ++i)
        q1[tim].push(x[i]);
    return MP(rt, edg);
}

int top(int k){
    while(1){
        if(q1[k].empty()) return -inf;
        if(q2[k].empty()) return q1[k].top();
        if(q1[k].top() == q2[k].top())
            q1[k].pop(), q2[k].pop();
        else return q1[k].top();
    }
}
void solve(int u){
    ++tim;
    pii ee = find_edge(u);
    if(ee.first == -1){
        ans[tim] = 0; return ;
    }
    int i = ee.second, u1 = vv[i], u2 = vv[i^1];
    vis[i] = vis[i^1] = 1;
    int cur = tim;
    lc[cur] = tim + 1;
    solve(u1);
    rc[cur] = tim + 1;
    solve(u2);
    ans[cur] = top(lc[cur]) + top(rc[cur]) + C[cur];
    ans[cur] = max(ans[cur], ans[lc[cur]]);
    ans[cur] = max(ans[cur], ans[rc[cur]]);
}
void gao(int u){
    for(int i = f[u]; ~i; i = nxt[i]){
        int v = vv[i], d = cost[i];
        if(col[u]) q1[v].push(d);
        else q2[v].push(d);
        if(lc[v]){
            ans[v] = top(lc[v]) + top(rc[v]) + C[v];
            ans[v] = max(ans[v], ans[lc[v]]);
            ans[v] = max(ans[v], ans[rc[v]]);
        }
        else{
            if(col[u]) ans[v] = 0;
            else ans[v] = -inf;
        }
    }
}
int main(){
    int n;
    scanf("%d", &n);
    init();
    for(int i = 1; i < n; ++i){
        int u, v, c;
        scanf("%d%d%d", &u, &v, &c);
        _add(u, v, c), _add(v, u, c);
    }
    tim = 0;
    solve(1);
    int q, white = n;
    scanf("%d", &q);
    while(q--){
        char s[5];
        scanf("%s", s);
        if(s[0] == 'A'){
            if(white == 0)
                puts("They have disappeared.");
            else printf("%d\n", ans[1]);
        }
        else{
            int u;
            scanf("%d", &u);

            gao(u);

            if(col[u]) white++;
            else white--;
            col[u] ^= 1;
        }
    }
    return 0;
}