BZOJ2243 [SDOI2011]染色(树链剖分+线段树合并)

题目链接 BZOJ2243

树链剖分 ++ 线段树

线段树每个节点维护lclc, rcrc, ss

lclc代表该区间的最左端的颜色,rcrc代表该区间的最右端的颜色

ss代表该区间的所有连续颜色段数(仅考虑该区间)

lazylazy表示延迟信息。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
#include <bits/stdc++.h>
 
using namespace std;
 
#define rep(i, a, b)    for (int i(a); i <= (b); ++i)
#define dec(i, a, b)    for (int i(a); i >= (b); --i)
#define lson        i << 1, L, mid
#define rson        i << 1 | 1, mid + 1, R
 
typedef long long LL;
 
const int N = 1e5 + 10;
const int A = 21;
 
vector <int> v[N];
int n, m;
int c[N];
int f[N][A];
int lazy[N << 2], s[N << 2], lc[N << 2], rc[N << 2];
int h[N], deep[N], sz[N];
int son[N];
int tot = 0;
int top[N];
 
 
int LCA(int a, int b){
    if (deep[a] < deep[b]) swap(a, b);
    for (int i = 0,  delta = deep[a] - deep[b]; delta; delta >>= 1, ++i) if (delta & 1) a = f[a][i];
    if (a == b) return a;
    dec(i, 19, 0) if (f[a][i] != f[b][i]) a = f[a][i], b = f[b][i];
    return f[a][0];
}
 
inline void pushup(int i){
    lc[i] = lc[i << 1];
    rc[i] = rc[i << 1 | 1];
    if (rc[i << 1] ^ lc[i << 1 | 1]) s[i] = s[i << 1] + s[i << 1 | 1];
    else s[i] = s[i << 1] + s[i << 1 | 1] - 1;
}
 
inline void pushdown(int i, int L, int R){
    int tmp = lazy[i];
    if (tmp == -1 || L == R) return;
 
    s[i << 1]      = s[i << 1 | 1]    = 1;
    lazy[i << 1]   = lazy[i << 1 | 1] = tmp;
 
    lc[i << 1]     = rc[i << 1]       = tmp;
    lc[i << 1 | 1] = rc[i << 1 | 1]   = tmp;
    lazy[i] = -1;
}  
 
void dfs1(int x, int fa, int dep){
    sz[x] = 1;
    deep[x] = dep;
 
    if (fa){
                f[x][0] = fa;
        for (int i = 0; f[f[x][i]][i]; ++i) f[x][i + 1] = f[f[x][i]][i];
    }
 
    int ct = (int)v[x].size();
 
    rep(i, 0, ct - 1){
        int u = v[x][i];
        if (u == fa) continue;
        dfs1(u, x, dep + 1);
        sz[x] += sz[u];
        if (sz[son[x]] < sz[u]) son[x] = u;
    }
}
 
void dfs2(int x, int tp){
    h[x] = ++tot;
    top[x] = tp;
    if (son[x]) dfs2(son[x], tp);
     
    int ct = (int)v[x].size();
    rep(i, 0, ct - 1){
        int u = v[x][i];
        if (u == f[x][0] || u == son[x]) continue;
        dfs2(u, u);
    }
}
 
void build(int i, int L, int R){
    s[i] = 1;
    lazy[i] = -1;
 
    if (L == R) return;
    int mid = (L + R) >> 1;
 
    build(lson);
    build(rson);
}
 
void change(int i, int L, int R, int l, int r, int val){
    pushdown(i, L, R);
    if (L == l && R == r){
        lc[i] = rc[i] = val;
        s[i]  = 1;
        lazy[i] = val;
        return;
    }
 
    int mid = (L + R) >> 1;
    if (r <= mid) change(lson, l, r, val);
    else if (l > mid) change(rson, l, r, val);
    else{
        change(lson, l, mid, val);
        change(rson, mid + 1, r, val);
    }
 
    pushup(i);
}
 
int query(int i, int L, int R, int l, int r){
    pushdown(i, L, R);
    if (L == l && R == r) return s[i];
 
    int mid = (L + R) >> 1;
    if (r <= mid) return query(lson, l, r);
    else if (l > mid) return query(rson, l, r);
    else{
        int tmp = 1;
        if (rc[i << 1] ^ lc[i << 1 | 1]) tmp = 0;
        return query(lson, l, mid) + query(rson, mid + 1, r) - tmp;
    }
}
 
int getcolor(int i, int L, int R, int x){
    pushdown(i, L, R);
    if (L == R) return lc[i];
    int mid = (L + R) >> 1;
    if (x <= mid) return getcolor(lson, x);
    else return getcolor(rson, x);
}
 
int solvesum(int x, int tp){
    int ret = 0;
    for (; top[x] ^ top[tp] ;){
        ret += query(1, 1, n, h[top[x]], h[x]);
        if (getcolor(1, 1, n, h[top[x]]) == getcolor(1, 1, n, h[f[top[x]][0]])) --ret;
        x = f[top[x]][0];
    }
 
    ret += query(1, 1, n, h[tp], h[x]);
    return ret;
}
 
void solvechange(int x, int tp, int val){
    for (; top[x] ^ top[tp]; ){
        change(1, 1, n, h[top[x]], h[x], val);
        x = f[top[x]][0];
    }
 
    change(1, 1, n, h[tp], h[x], val);
}
 
 
void solve(){
    int x, y, z;
    dfs1(1, 0, 0);
    dfs2(1, 1);
    build(1, 1, n);
    rep(i, 1, n) change(1, 1, n, h[i], h[i], c[i]);
 
    rep(i, 1, m){
        char ch[10];
        scanf("%s", ch);
        if (ch[0] == 'Q'){
            scanf("%d%d", &x, &y);
            int t = LCA(x, y);
            printf("%d\n", solvesum(x, t) + solvesum(y, t) - 1);
        }
        else{
            scanf("%d%d%d", &x, &y, &z);
            int t = LCA(x, y);
            solvechange(x, t, z);
            solvechange(y, t, z);
        }
    }
}
 
void init(){
    scanf("%d%d", &n, &m);
    rep(i, 1, n) scanf("%d", c + i);
    rep(i, 2, n){
        int x, y;
        scanf("%d%d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
}
 
 
int main(){
    init();
    solve();
    return 0;
}

 

posted @   cxhscst2  阅读(186)  评论(0编辑  收藏  举报
努力加载评论中...
点击右上角即可分享
微信分享提示