Bzoj1558: [JSOI2009]等差数列

题面

传送门

Sol

线段树维护区间$DP$
差分，你会发现就是选一些区间，第一个值可以不一样
那么我们维护原数组左右端点是否选的情况，一共四种
注意差分数组只有$n-1$的长度，并且每个数维护的是两个相邻的原数组的数

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
using namespace std;
typedef long long ll;
const int _(1e5 + 5);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    return x * z;
}

int n, Q, a[_];
struct Data{
    int s[4], l, r;
    //0:l0r0, 1:l1r0, 2:l0r1, 3:l1r1
};
struct Segment{
    int tag;
    Data v;
} T[_ << 2];

IL Data Merge(RG Data A, RG Data B){
    RG Data C; RG int tmp = A.r == B.l;
    C.l = A.l, C.r = B.r;
    C.s[0] = A.s[2] + B.s[1] - tmp;
    C.s[0] = min(C.s[0], min(A.s[0] + B.s[1], A.s[2] + B.s[0]));
    C.s[1] = A.s[3] + B.s[1] - tmp;
    C.s[1] = min(C.s[1], min(A.s[1] + B.s[1], A.s[3] + B.s[0]));
    C.s[2] = A.s[2] + B.s[3] - tmp;
    C.s[2] = min(C.s[2], min(A.s[0] + B.s[3], A.s[2] + B.s[2]));
    C.s[3] = A.s[3] + B.s[3] - tmp;
    C.s[3] = min(C.s[3], min(A.s[1] + B.s[3], A.s[3] + B.s[2]));
    return C;
}

IL void Build(RG int x, RG int l, RG int r){
    RG Data &X = T[x].v;
    if(l == r){
        X.s[1] = X.s[2] = X.s[3] = 1, X.l = X.r = a[l] - a[l - 1];
        return;
    }
    RG int mid = (l + r) >> 1;
    Build(x << 1, l, mid), Build(x << 1 | 1, mid + 1, r);
    X = Merge(T[x << 1].v, T[x << 1 | 1].v);
}

IL void Adjust(RG int x, RG int tag){
    RG Data &X = T[x].v;
    T[x].tag += tag, X.l += tag, X.r += tag;
}

IL void Pushdown(RG int x){
    if(!T[x].tag) return;
    Adjust(x << 1, T[x].tag), Adjust(x << 1 | 1, T[x].tag);
    T[x].tag = 0;
}

IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){
    if(L <= l && R >= r){
        Adjust(x, v);
        return;
    }
    Pushdown(x);
    RG int mid = (l + r) >> 1;
    if(L <= mid) Modify(x << 1, l, mid, L, R, v);
    if(R > mid) Modify(x << 1 | 1, mid + 1, r, L, R, v);
    T[x].v = Merge(T[x << 1].v, T[x << 1 | 1].v);
}

IL Data Query(RG int x, RG int l, RG int r, RG int L, RG int R){
    if(L == l && R == r) return T[x].v;
    Pushdown(x);
    RG int mid = (l + r) >> 1;
    if(R <= mid) return Query(x << 1, l, mid, L, R);
    else if(L > mid) return Query(x << 1 | 1, mid + 1, r, L, R);
    else return Merge(Query(x << 1, l, mid, L, mid), Query(x << 1 | 1, mid + 1, r, mid + 1, R));
}

int main(RG int argc, RG char *argv[]){
    n = Input();
    for(RG int i = 0; i < n; ++i) a[i] = Input();
    --n, Build(1, 1, n);
    for(Q = Input(); Q; --Q){
        RG char op; scanf(" %c", &op);
        if(op == 'B'){
            RG int l = Input(), r = Input();
            (l == r) ? puts("1") : printf("%d\n", Query(1, 1, n, l, r - 1).s[3]);
        }
        else{
            RG int l = Input(), r = Input(), x = Input(), y = Input();
            if(l > 1) Modify(1, 1, n, l - 1, l - 1, x);
            if(l < r) Modify(1, 1, n, l, r - 1, y);
            if(r < n) Modify(1, 1, n, r, r, -(x + y * (r - l)));
        }
    }
    return 0;
}