[SCOI2010]序列操作 BZOJ1858 線段樹
阿新 • • 發佈：2018-12-14
dll san true 序列 lse 端點 是否 dst ==
題目描述
lxhgww最近收到了一個01序列，序列裏面包含了n個數，這些數要麽是0，要麽是1，現在對於這個序列有五種變換操作和詢問操作：
0 a b 把[a, b]區間內的所有數全變成0
1 a b 把[a, b]區間內的所有數全變成1
2 a b 把[a,b]區間內的所有數全部取反，也就是說把所有的0變成1，把所有的1變成0
3 a b 詢問[a, b]區間內總共有多少個1
4 a b 詢問[a, b]區間內最多有多少個連續的1
對於每一種詢問操作，lxhgww都需要給出回答，聰明的程序員們，你們能幫助他嗎？
輸入輸出格式
輸入格式：輸入數據第一行包括2個數，n和m，分別表示序列的長度和操作數目
第二行包括n個數，表示序列的初始狀態
接下來m行，每行3個數，op, a, b，（0<=op<=4，0<=a<=b<n）表示對於區間[a, b]執行標號為op的操作
輸出格式：對於每一個詢問操作，輸出一行，包括1個數，表示其對應的答案
輸入輸出樣例
輸入樣例#1： 復制10 10 0 0 0 1 1 0 1 0 1 1 1 0 2 3 0 5 2 2 2 4 0 4 0 3 6 2 3 7 4 2 8 1 0 5 0 5 6 3 3 9輸出樣例#1： 復制
5 2 6 5
說明
對於30%的數據，1<=n, m<=1000
對於100%的數據，1<=n, m<=100000
很刺激的線段樹題目；
註意賦值的操作的優先級>翻轉的操作；
還有很多細節要註意；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
struct node {
int sum;
int Lw, Lb;// Lb:區間連續0的最大長度，Lw:連續1的最大長度
int llw, rlw;//llb:區間左端點連續0的最大長度，llw:區間左端點連續1的最大長度；
int llb, rlb;// rlb:區間右端點連續0的最大長度；rlw：區間右端點連續1的最大長度
int lazy, rev;// lazy：是否全部變為1/0；rev：翻轉標記
}tr[maxn<<2];
int n, m;
#define lson (rt<<1)
#define rson (rt<<1|1)
#define mid ((l+r)>>1)
void pushup(int l, int r, int rt) {
tr[rt].sum = tr[lson].sum + tr[rson].sum;
tr[rt].Lb = max(tr[lson].Lb, tr[rson].Lb);
tr[rt].Lw = max(tr[lson].Lw, tr[rson].Lw);
tr[rt].llb = tr[lson].llb;
tr[rt].llw = tr[lson].llw;
if (tr[lson].llw == mid - l + 1)tr[rt].llw += tr[rson].llw;
if (tr[lson].llb == mid - l + 1)tr[rt].llb += tr[rson].llb;
tr[rt].rlb = tr[rson].rlb;
tr[rt].rlw = tr[rson].rlw;
if (tr[rson].rlb == r - mid)tr[rt].rlb += tr[lson].rlb;
if (tr[rson].rlw == r - mid)tr[rt].rlw += tr[lson].rlw;
tr[rt].Lw = max(tr[rt].Lw, tr[lson].rlw + tr[rson].llw);
tr[rt].Lb = max(tr[rt].Lb, tr[lson].rlb + tr[rson].llb);
}
void down(int l, int r, int rt) {
int lazy = tr[rt].lazy, rev = tr[rt].rev;
int L1 = mid - l + 1, L2 = r - mid;
tr[rt].lazy = -1;
if (lazy == 0) {
tr[lson] = node { 0, 0, L1, 0, 0, L1, L1, 0, 0 };
tr[rson] = node{ 0,0,L2,0,0,L2,L2,0,0 };
}
else if (lazy == 1) {
tr[lson] = node{ L1,L1,0,L1,L1,0,0,1,0 };
tr[rson] = node{ L2,L2,0,L2,L2,0,0,1,0 };
}
if (rev) {
tr[rt].rev = 0;
int sum = tr[lson].sum;
int Lw = tr[lson].Lw, Lb = tr[lson].Lb;
int llw = tr[lson].llw, llb = tr[lson].llb;
int rlw = tr[lson].rlw, rlb = tr[lson].rlb;
tr[lson].sum = L1 - sum;
tr[lson].Lw = Lb; tr[lson].Lb = Lw;
tr[lson].llw = llb; tr[lson].rlw = rlb;
tr[lson].llb = llw; tr[lson].rlb = rlw;
tr[lson].rev ^= 1;
sum = tr[rson].sum;
Lw = tr[rson].Lw; Lb = tr[rson].Lb;
llw = tr[rson].llw; llb = tr[rson].llb;
rlw = tr[rson].rlw; rlb = tr[rson].rlb;
tr[rson].sum = L2 - sum;
tr[rson].Lw = Lb; tr[rson].Lb = Lw;
tr[rson].llw = llb; tr[rson].rlw = rlb;
tr[rson].llb = llw; tr[rson].rlb = rlw;
tr[rson].rev ^= 1;
}
}
void build(int l, int r, int rt) {
tr[rt].lazy = -1;
if (l == r) {
int x; rdint(x);
tr[rt] = node{ x,x,(x ^ 1),x,x,(x ^ 1),(x ^ 1),-1,0 };
return;
}
build(l, mid, lson); build(mid + 1, r, rson);
pushup(l, r, rt);
}
void upd1(int l, int r, int rt, int L, int R) {
if (L <= l && r <= R) {
int LL = r - l + 1;
tr[rt] = node{ 0,0,LL,0,0,LL,LL,0,0 };
return;
}
if (l > R || r < L)return;
down(l, r, rt);
upd1(l, mid, lson, L, R); upd1(mid + 1, r, rson, L, R);
pushup(l, r, rt);
}
void upd2(int l, int r, int rt, int L, int R) {
if (L <= l && r <= R) {
int LL = r - l + 1;
tr[rt] = node{ LL,LL,0,LL,LL,0,0,1,0 };
return;
}
if (l > R || r < L)return;
down(l, r, rt);
upd2(l, mid, lson, L, R); upd2(mid + 1, r, rson, L, R);
pushup(l, r, rt);
}
void upd3(int l, int r, int rt, int L, int R) {
if (L <= l && r <= R) {
int LL = r - l + 1, sum = tr[rt].sum, Lw = tr[rt].Lw, Lb = tr[rt].Lb;
int llw = tr[rt].llw, llb = tr[rt].llb;
int rlw = tr[rt].rlw, rlb = tr[rt].rlb;
tr[rt] = node{ LL - sum,Lb,Lw,llb,rlb,llw,rlw,tr[rt].lazy,(tr[rt].rev ^ 1) };
return;
}
if (l > R || r < L)return;
down(l, r, rt);
upd3(l, mid, lson, L, R);
upd3(mid + 1, r, rson, L, R);
pushup(l, r, rt);
}
int query1(int l, int r, int rt, int L, int R) {
if (L <= l && r <= R)return tr[rt].sum;
if (l > R || r < L)return 0;
down(l, r, rt);
return query1(l, mid, lson, L, R) + query1(mid + 1, r, rson, L, R);
}
node query2(int l, int r, int rt, int L, int R) {
if (L <= l && r <= R)return tr[rt];
if (l > R || r < L)return node{ 0 };
down(l, r, rt);
node LL = query2(l, mid, lson, L, R), RR = query2(mid + 1, r, rson, L, R), ans;
ans.sum = LL.sum + RR.sum;
ans.Lw = max(LL.Lw, RR.Lw);
ans.Lb = max(LL.Lb, RR.Lb);
ans.llw = LL.llw, ans.rlw = RR.rlw;
ans.llb = LL.llb, ans.rlb = RR.rlb;
if (LL.llb == mid - l + 1)ans.llb += RR.llb;
if (LL.llw == mid - l + 1)ans.llw += RR.llw;
if (RR.rlb == r - mid)ans.rlb += LL.rlb;
if (RR.rlw == r - mid)ans.rlw += LL.rlw;
ans.Lw = max(ans.Lw, LL.rlw + RR.llw);
ans.Lb = max(ans.Lb, LL.rlb + RR.llb);
return ans;
}
int main()
{
//ios::sync_with_stdio(0);
rdint(n); rdint(m);
build(1, n, 1);
while (m--) {
int op, a, b;
rdint(op); rdint(a); rdint(b);
a++; b++;
if (op == 0)upd1(1, n, 1, a, b);
if (op == 1)upd2(1, n, 1, a, b);
if (op == 2)upd3(1, n, 1, a, b);
if (op == 3)cout << query1(1, n, 1, a, b) << endl;
if (op == 4)cout << query2(1, n, 1, a, b).Lw << endl;
}
return 0;
}
[SCOI2010]序列操作 BZOJ1858 線段樹