牛客竞赛_ACM/NOI/CSP/CCPC/ICPC算法编程高难度练习赛_牛客竞赛OJ
G 智乃酱的平方数列（线段树，等差数列，多项式）
题目描述
想必你一定会用线段树维护等差数列吧？让我们来看看它的升级版。
请你维护一个长度为5×10 ^5的数组，一开始数组中每个元素都为0，要求支持以下两个操作：
1、区间[l,r]加自然数的平方数组，即al+=1，al+1+=4，al+2+=9，al+3+=16...ar+=(r−l+1)∗(r−l+1)
2、区间[l,r]查询区间和mod 10^9 + 7
输入描述:
第一行输入n,m,(1≤n,m≤5*10 ^5)
接下来m行，对于每行，先读入一个整数q。
当q的值为1时，还需读入两个整l,r,(1≤l≤r≤n)表示需要对区间[l,r]进行操作，让第一个元素加1，第二个元素加4，第三个元素加9...以此类推。
当q的值为2时，还需读入两个整数l,r(1≤l≤r≤n)表示查询l到r的元素和
输出描述:
对于每一个q=2，输出一行一个非负整数，表示l到r的区间和mod 110^9+7。
示例1
输入
复制
4 4
2 1 4
1 1 4
1 3 4
2 1 4
输出
复制
0
35
示例2
输入
复制
10 6
1 1 6
1 8 9
1 3 6
2 1 10
1 1 10
2 1 10
输出
复制
126
511
解析：
等差数列可以写成 [x-(l-1)]^2,其中 x 表示当前的位置，等价于 x^2-2*x*(l-1)+(l-1)^2，通过线段树维护其系数即可
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <utility>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <math.h>
#include <map>
#include <sstream>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <stdio.h>
#include <tuple>
using namespace std;
/*
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
*/
typedef long long LL;
#define int long long
#define ld long double
//#define INT __int128
const LL INF = 0x3f3f3f3f3f3f3f3f;
typedef unsigned long long ULL;
typedef pair<long long, long long> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int inf = 0x3f3f3f3f;
const LL mod = 1e9+7;
const ld eps = 1e-12;
const int N = 5e5 + 10, M = N + 10;
int n, m;
struct TREE {int l, r;int la1, la2, la3;int sum;int base1, base2;
}tr[N << 2];
#define ls u<<1
#define rs u<<1|1
void cal(int u, int la1, int la2, int la3) {la3 %= mod;int len = tr[u].r - tr[u].l + 1;tr[u].sum = (tr[u].sum + len * la3 % mod) % mod;tr[u].sum = (tr[u].sum - 2 * tr[u].base2 * la2 % mod + mod) % mod;tr[u].sum = (tr[u].sum + tr[u].base1 * la1 % mod) % mod;tr[u].la1 = (tr[u].la1 + la1) % mod;tr[u].la2 = (tr[u].la2 + la2) % mod;tr[u].la3 = (tr[u].la3 + la3) % mod;
}
void up(int u) {tr[u].sum = (tr[ls].sum + tr[rs].sum) % mod;
}
void down(int u) {if (tr[u].la1||tr[u].la2||tr[u].la3) {cal(ls, tr[u].la1, tr[u].la2, tr[u].la3);cal(rs, tr[u].la1, tr[u].la2, tr[u].la3);tr[u].la1 = tr[u].la2 = tr[u].la3 = 0;}
}
void build(int u, int l, int r) {tr[u].l = l, tr[u].r = r;if (l == r) {tr[u].base1 = (l * l) % mod , tr[u].base2 = l;return;}int mid = (l + r) >> 1;build(ls, l, mid), build(rs, mid + 1, r);tr[u].base1 = (tr[ls].base1 + tr[rs].base1) % mod;tr[u].base2 = (tr[ls].base2 + tr[rs].base2) % mod;
}
void modify(int u, int l, int r, int d) {if (l <= tr[u].l && tr[u].r <= r) {cal(u, 1, d, d * d);return;}down(u);int mid = (tr[u].l + tr[u].r) >> 1;if (l <= mid)modify(ls, l, r, d);if (r > mid)modify(rs, l, r, d);up(u);
}
int query(int u, int l, int r) {if (l <= tr[u].l && tr[u].r <= r) {return tr[u].sum;}down(u);int ret = 0;int mid = (tr[u].l + tr[u].r) >> 1;if (l <= mid)ret = (ret + query(ls, l, r)) % mod;if (r > mid)ret = (ret + query(rs, l, r)) % mod;//up(u);return ret;
}
signed main() {ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);cin >> n >> m;build(1, 1, n);int op, l, r;while (m--) {cin >> op >> l >> r;if (op == 1) {modify(1, l, r, l - 1);}else {cout << query(1, l, r) << endl;}}return 0;
}
/*
4 5
2 1 4
1 1 4
2 1 4
1 3 4
2 1 4
4 5
2 1 4
1 3 4
2 1 4
1 1 4
2 1 4
*/K 智乃酱的双塔问题·改（线段树，dp，矩阵连乘)
解析：
不能发现这个问题如果不带修改则可以用 dp+前缀和（矩阵乘法）解决，带修改则可以用线段树维护区间和。
令 f[i][0/1] 表示从第 1 层的左边（0）或右边（1）到第 i 层的左边（0）或右边（1）的方案数
我们可以将每一层的转移状态用矩阵表示，
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <utility>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <math.h>
#include <map>
#include <sstream>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <stdio.h>
#include <tuple>
using namespace std;
/*
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
*/
typedef long long LL;
//#define int long long
#define ld long double
//#define INT __int128
const LL INF = 0x3f3f3f3f3f3f3f3f;
typedef unsigned long long ULL;
typedef pair<long long, long long> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int inf = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const ld eps = 1e-12;
const int N = 2e5 + 10, M = N + 10;
int n, m;
char s[N];
struct MATRIX {LL a[2][2];void init(int b1,int b2,int b3,int b4) {a[0][0] = b1, a[0][1] = b2;a[1][0] = b3, a[1][1] = b4;}MATRIX operator*(MATRIX oth) {MATRIX ret;ret.init(0, 0, 0, 0);for (int i = 0; i < 2; i++) {for (int j = 0; j < 2; j++) {for (int k = 0; k < 2; k++) {(ret.a[i][j] += a[i][k] * oth.a[k][j] % mod) %= mod;}}}return ret;}
};struct TREE {int l, r;MATRIX ma;
}tr[N << 2];
#define ls u<<1
#define rs u<<1|1
void up(int u) {tr[u].ma = tr[ls].ma * tr[rs].ma;
}
void build(int u, int l, int r) {tr[u].l = l, tr[u].r = r;if (l == r) {tr[u].ma.init(1, s[l] == '/', s[l] != '/', 1);return;}int mid = l + r >> 1;build(ls, l, mid), build(rs, mid + 1, r);up(u);
}
void modify(int u, int pos, char ch) {if (tr[u].l == tr[u].r) {tr[u].ma.init(1, ch == '/', ch != '/', 1);return;}int mid = tr[u].l + tr[u].r >> 1;if (pos <= mid)modify(ls, pos, ch);if (pos > mid)modify(rs, pos, ch);up(u);
}
MATRIX query(int u, int l, int r) {if (l <= tr[u].l && tr[u].r <= r) {return tr[u].ma;}int mid = tr[u].l + tr[u].r >> 1;MATRIX ret;ret.init(1, 0, 0, 1);if (l <= mid)ret = (ret * query(ls, l, r));if (r > mid)ret = (ret * query(rs, l, r));return ret;
}
signed main() {cin >> n >> m;scanf("%s", s + 1);build(1, 1, n - 1);int op, ht, hs, h;for (int i = 1; i <= m; i++) {scanf("%d", &op);if (op == 0) {char c[2];scanf("%d%s",&h, c);modify(1, h,c[0]);}else {int ps, pt;scanf("%d%d%d%d",&hs,&ht, &ps, &pt);auto ret = query(1, hs, ht - 1);/*cout << "__________________" << endl;for (int i = 0; i < 2; i++) {cout << ret.a[i][0] << " " << ret.a[i][1] << endl;}cout << endl;*/LL ans = ret.a[ps][pt];printf("%lld\n", ans);}}return 0;
}J 智乃酱的双塔问题·极（带修改的DP，DDP）
这题思路与上题一样，只不过将矩阵的乘法运算换成 floyd 算法，矩阵的含义变为图的邻接矩阵。（可以变的主要原因是因为 floyd 算法的运算形式与矩阵连乘的形式很相似）
#include <iostream>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <utility>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <math.h>
#include <map>
#include <sstream>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <stdio.h>
#include <tuple>
using namespace std;
/*
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
*/
typedef long long LL;
//#define int long long
#define ld long double
//#define INT __int128
const LL INF = 0x3f3f3f3f3f3f3f3f;
typedef unsigned long long ULL;
typedef pair<long long, long long> PLL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int inf = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const ld eps = 1e-12;
const int N = 1e5 + 10, M = N + 10;
int n, m;
int val[N][3];
char s[N];
#define inf INF
struct MATRIX {LL a[2][2];void init(LL b1, LL b2, LL b3, LL b4) {a[0][0] = b1, a[0][1] = b2;a[1][0] = b3, a[1][1] = b4;}MATRIX operator*(MATRIX oth) {MATRIX ret;ret.init(inf, inf, inf, inf);for (int i = 0; i < 2; i++) {for (int j = 0; j < 2; j++) {for (int k = 0; k < 2; k++) {//(ret.a[i][j] += a[i][k] * oth.a[k][j] % mod) %= mod;if (a[i][k] == inf || oth.a[k][j] == inf)continue;ret.a[i][j] = min(ret.a[i][j], (a[i][k] + oth.a[k][j]));}}}return ret;}
};struct TREE {int l, r;MATRIX ma;
}tr[N << 2];
#define ls u<<1
#define rs u<<1|1
void up(int u) {tr[u].ma = tr[ls].ma * tr[rs].ma;
}
void build(int u, int l, int r) {tr[u].l = l, tr[u].r = r;if (l == r) {if (s[l] == '/')tr[u].ma.init(val[l][0], val[l][2], inf, val[l][1]);else tr[u].ma.init(val[l][0], inf, val[l][2], val[l][1]);return;}int mid = l + r >> 1;build(ls, l, mid), build(rs, mid + 1, r);up(u);
}
void modify(int u, int pos, char ch, int val1, int val2, int val3) {if (tr[u].l == tr[u].r) {if (ch == '/')tr[u].ma.init(val1, val3, inf, val2);else tr[u].ma.init(val1, inf, val3, val2);return;}int mid = tr[u].l + tr[u].r >> 1;if (pos <= mid)modify(ls, pos, ch, val1, val2, val3);if (pos > mid)modify(rs, pos, ch, val1, val2, val3);up(u);
}
MATRIX query(int u, int l, int r) {if (l <= tr[u].l && tr[u].r <= r) {return tr[u].ma;}int mid = tr[u].l + tr[u].r >> 1;MATRIX ret;ret.init(0, inf, inf, 0);if (l <= mid)ret = ret * query(ls, l, r);if (r > mid)ret = ret * query(rs, l, r);return ret;
}
signed main() {cin >> n >> m;scanf("%s", s + 1);for (int i = 1; i < n; i++) {scanf("%d%d%d", &val[i][0], &val[i][1], &val[i][2]);}build(1, 1, n - 1);int op, ht, hs, h;for (int i = 1; i <= m; i++) {scanf("%d", &op);if (op == 0) {char c[2];scanf("%d%s", &h, c);//cout << "______" << c[0] << endl;s[h] = c[0];modify(1, h, c[0], val[h][0], val[h][1], val[h][2]);}else if (op == 1) {int val1, val2, val3;scanf("%d%d%d%d", &h, &val1, &val2, &val3);val[h][0] = val1, val[h][1] = val2, val[h][2] = val3;modify(1, h, s[h], val1, val2, val3);}else {int ps, pt;scanf("%d%d%d%d", &hs, &ht, &ps, &pt);auto ret = query(1, hs, ht - 1);/*cout << "__________________" << endl;for (int i = 0; i < 2; i++) {cout << ret.a[i][0] << " " << ret.a[i][1] << endl;}cout << endl;*/LL ans = ret.a[ps][pt];if (ans == inf) {printf("-1\n");}elseprintf("%lld\n", ans);}}return 0;
}/*
4 13
///
1 2 1
2 3 5
8 8 1
2 1 4 1 0
2 1 2 1 0
2 2 3 1 0
2 3 4 1 0
0 3 \
2 1 4 0 0
2 2 4 0 0
2 2 4 0 1
2 3 4 0 1
2 3 4 1 0
1 1 1 1 1
1 2 1 1 1
1 3 1 1 1
2 1 4 1 0
0 3 /
2 1 4 1 04 13
///
1 2 1
2 3 5
8 8 1
2 1 4 1 0
0 3 \
2 1 4 0 0
2 2 4 0 0
2 2 4 0 1
2 3 4 0 1
2 3 4 1 0
1 1 1 1 1
1 2 1 1 1
1 3 1 1 1
2 1 4 1 0
0 3 /
2 1 4 1 0
*/
