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HYSBZ 2243 染色 (线段树+树链剖分)

2017-09-05 21:03 429 查看
题意:中文题。

析:真是一个好题,但是我TLE了两天,就是因为输入那个询问数,我当作边数了,结果就是一个TLE。。。

大体思路,就是先进行用树链剖分,然后用线段树来维护,维护每个区间的不同数的个数,和每个数的值,在求的时候,在两个端点进行判断,是不是同一种,如果是就减去1,不是则不变。

而且发现一个问题,就是网上的代码所以输出的和我的不一样,但是都AC了,不知道是不是数据水。也不知道谁的对。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 50;
const int mod = 1000;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r > 0 && r <= n && c > 0 && c <= m;
}

struct Edge{
int to, next;
};
Edge edge[maxn<<1];
int head[maxn], tot, top[maxn], dep[maxn], a[maxn], fp[maxn];
int son[maxn], fa[maxn], num[maxn], p[maxn], pos;
int sum[maxn<<2], setv[maxn<<2], lx[maxn<<2], rx[maxn<<2], value[maxn<<2];

void init(){
tot = pos = 0;
ms(head, -1);  ms(son, -1);
ms(setv, -1);
}

void addEdge(int u, int v){
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}

void dfs1(int u, int f, int d){
dep[u] = d;  fa[u] = f;
num[u] = 1;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == f)  continue;
dfs1(v, u, d+1);
num[u] += num[v];
if(son[u] == -1 || num[son[u]] < num[v])  son[u] = v;
}
}

void dfs2(int u, int sp){
top[u] = sp;  p[u] = ++pos;
fp[pos] = u;
if(son[u] == -1)  return ;
dfs2(son[u], sp);
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == son[u] || v == fa[u])  continue;
dfs2(v, v);
}
}

void push_up(int rt){
int l = rt<<1, r = rt<<1|1;
sum[rt] = sum[l] + sum[r];
if(rx[l] == lx[r])  --sum[rt];
lx[rt] = lx[l];  rx[rt] = rx[r];
}

void build(int l, int r, int rt){
if(l == r){
value[rt] = a[fp[l]];
sum[rt] = 1;
lx[rt] = rx[rt] = value[rt];
return ;
}
int m = l + r >> 1;
build(lson);
build(rson);
pu(rt);
}

void push_down(int rt){
if(setv[rt] < 0)  return ;
int l = rt<<1, r = rt<<1|1;
sum[l] = sum[r] = 1;
value[l] = value[r] = setv[rt];
lx[l] = rx[l] = lx[r] = rx[r] = setv[rt];
setv[l] = setv[r] = setv[rt];
setv[rt] = -1;
}

void update(int L, int R, int val, int l, int r, int rt){
if(L <= l && r <= R){
sum[rt] = 1;
value[rt] = setv[rt] = val;
lx[rt] = rx[rt] = val;
return ;
}
pd(rt);
int m = l + r >> 1;
if(L <= m)  update(L, R, val, lson);
if(R > m)   update(L, R, val, rson);
pu(rt);
}

int queryVal(int M, int l, int r, int rt){
if(l == r)  return value[rt];
pd(rt);
int m = l + r >> 1;
if(M <= m)  return queryVal(M, lson);
return queryVal(M, rson);
}

int query(int L, int R, int l, int r, int rt){
if(L <= l && r <= R)  return sum[rt];
pd(rt);
int m = l + r >> 1;
if(L <= m && R > m){
int ans = query(L, R, lson) + query(L, R, rson);
if(rx[rt<<1] == lx[rt<<1|1])  --ans;
return ans;
}
else if(L <= m)  return query(L, R, lson);
else  return query(L, R, rson);
}

void update(int u, int v, int c){
int f1 = top[u], f2 = top[v];
while(f1 != f2){
if(dep[f1] < dep[f2]){
swap(f1, f2);
swap(u, v);
}
update(p[f1], p[u], c, all);
u = fa[f1];
f1 = top[u];
}
if(dep[u] > dep[v])  swap(u, v);
update(p[u], p[v], c, all);
}

int query(int u, int v){
int f1 = top[u], f2 = top[v];
int ans = 0;
while(f1 != f2){
if(dep[f1] < dep[f2]){
swap(u, v);
swap(f1, f2);
}
ans += query(p[f1], p[u], all);
if(queryVal(p[f1], all) == queryVal(p[fa[f1]], all))  --ans;
u = fa[f1];
f1 = top[u];
}
if(dep[u] > dep[v])  swap(u, v);
ans += query(p[u], p[v], all);
return ans;
}

int main(){
scanf("%d %d", &n, &m);
init();
for(int i = 1; i <= n; ++i)  scanf("%d", a+i);
for(int i = 1; i < n; ++i){
int u, v;
scanf("%d %d", &u, &v);
addEdge(u, v);
addEdge(v, u);
}
dfs1(1, 0, 0);
dfs2(1, 1);
build(all);
char op[5];
while(m--){
scanf("%s", op);
int a, b, c;
scanf("%d %d", &a, &b);
if(op[0] == 'C'){
scanf("%d", &c);
update(a, b, c);
}
else  printf("%d\n", query(a, b));
}
return 0;
}


  
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