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洛谷P3313 [SDOI2014]旅行(树链剖分 动态开节点线段树)

2019年01月31日  | 移动技术网IT编程  | 我要评论

人猿泰山 成人,流动比率和速动比率,大雨磅礴

题意

题目链接

sol

树链剖分板子 + 动态开节点线段树板子

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
//#define int long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
const double eps = 1e-9;
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;}
template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int ch(int x, int l, int r) {
    return x >= l && x <= r;
}
int n, q, c[maxn], w[maxn], dep[maxn], fa[maxn], son[maxn], siz[maxn], top[maxn], id[maxn], cnt;
vector<int> v[maxn];
void dfs1(int x, int _fa) {
    dep[x] = dep[_fa] + 1; siz[x] = 1; fa[x] = _fa; 
    for(auto &to : v[x]) {
        if(to == _fa) continue;
        dfs1(to, x);
        siz[x] += siz[to];
        if(siz[to] > siz[son[x]]) son[x] = to;
    }
}
void dfs2(int x, int topf) {
    top[x] = topf; id[x] = ++cnt;
    if(!son[x]) return ;
    dfs2(son[x], topf);
    for(auto &to : v[x]) {
        if(top[to]) continue;
        dfs2(to, to);
    }
}
const int ss = (3e6 + 10);
int root[maxn], ls[ss], rs[ss], mx[ss], sum[ss], tot;
void update(int k) {
    mx[k] = max(mx[ls[k]], mx[rs[k]]);
    sum[k] = sum[ls[k]] + sum[rs[k]];
}
void modify(int &k, int l, int r, int p, int v) {
    if(!k) k = ++tot;
    if(l == r) {sum[k] = v, mx[k] = v; return ;}
    int mid = l + r >> 1;
    if(p <= mid) modify(ls[k], l, mid, p, v);
    if(p  > mid) modify(rs[k], mid + 1, r, p, v);
    update(k);
}
int query(int k, int l, int r, int ql, int qr, int opt) {
    if(ql <= l && r <= qr) return opt == 0 ? mx[k] : sum[k];
    int mid = l + r >> 1;
    if(ql > mid) return query(rs[k], mid + 1, r, ql, qr, opt);
    else if(qr <= mid) return query(ls[k], l, mid, ql, qr, opt);
    else return opt == 0 ? max(query(ls[k], l, mid, ql, qr, opt), query(rs[k], mid + 1, r, ql, qr, opt)) 
                         : query(ls[k], l, mid, ql, qr, opt) + query(rs[k], mid + 1, r, ql, qr, opt);
}
int treemax(int x, int y) {
    int ans = -inf, p = x;
    while(top[x] != top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        chmax(ans, query(root[c[p]], 1, n, id[top[x]], id[x], 0));
        x = fa[top[x]];
    }
    if(dep[x] < dep[y]) swap(x, y);
    chmax(ans, query(root[c[p]], 1, n, id[y], id[x], 0));
    return ans;
}
int treesum(int x, int y) {
    int ans = 0, p = x;
    while(top[x] != top[y]) {
        if(dep[top[x]] < dep[top[y]]) swap(x, y);
        ans += query(root[c[p]], 1, n, id[top[x]], id[x], 1);
        x = fa[top[x]];
    }
    if(dep[x] < dep[y]) swap(x, y);
    ans += query(root[c[p]], 1, n, id[y], id[x], 1);
    return ans;
}
signed main() {
    n = read(); q = read();
    for(int i = 1; i <= n; i++) w[i] = read(), c[i] = read();
    for(int i = 1; i <= n - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y);
        v[y].push_back(x);
    }
    dfs1(1, 0);
    dfs2(1, 1);
    for(int i = 1; i <= n; i++) 
        modify(root[c[i]], 1, n, id[i], w[i]);
    while(q--) {
        char s[4];
        scanf("%s", s);
        int x = read(), c = read();
        if(s[0] == 'c' && s[1] == 'c') {
            modify(root[c[x]], 1, n, id[x], 0);
            modify(root[c], 1, n, id[x], w[x]);
            c[x] = c;
        }
        if(s[0] == 'c' && s[1] == 'w') {
            modify(root[c[x]], 1, n, id[x], c);
            w[x] = c;
        }
        if(s[0] == 'q' && s[1] == 'm') {
            printf("%d\n", treemax(x, c));  
        }
        if(s[0] == 'q' && s[1] == 's') {
            printf("%d\n", treesum(x, c));
        }
    }
    return 0;
}
/*
5 6
3 1
2 3
1 2
3 3
5 1
1 2
1 3
3 4
3 5
qs 1 5
cc 3 1
qs 1 5
cw 3 3
qs 1 5
qm 2 4
*/

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