[Codeforces771E]Bear and Rectangle Strips

Problem

给你一个2*n的矩阵，要求你用补充叠的矩阵去框，要求每个矩阵框中的数之和为0，问最多可以用几个矩阵。

Solution

首先预处理出一个点到离它最近的一段和为0的区间的左端点
然后到这往前用记忆化搜索的方式DP就可以了

Notice

注意要记忆化

Code

#include<map>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define reg register int
#define rep(i, a, b) for (reg i = a; i <= b; i++)
#define per(i, a, b) for (reg i = a; i >= b; i--)
#define travel(i, u) for (reg i = head[u]; i; i = edge[i].next)
const int INF = 1e9, N = 300000;
const double eps = 1e-6, phi = acos(-1);
ll mod(ll a, ll b) {if (a >= b || a < 0) a %= b; if (a < 0) a += b; return a;}
ll read(){ ll x = 0; int zf = 1; char ch; while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar(); while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); return x * zf;}
void write(ll y) { if (y < 0) putchar('-'), y = -y; if (y > 9) write(y / 10); putchar(y % 10 + '0');}
map<ll, int> Derec;
map<pair<int, int>, int> F;
int Pre[3][N + 5];
ll T[3][N + 5];
int Solve(int n, int m)
{
	if (!~Pre[1][n] && !~Pre[2][m] && !~Pre[0][min(n, m)]) return 0;
	pair<int, int> P(n, m);
	if (F.count(P)) return F[P];
	int ans = 0;
	if (Pre[1][n] > Pre[2][m]) ans = max(ans, Solve(Pre[1][n], m) + 1);
	else if (~Pre[2][m])ans = max(ans, Solve(n, Pre[2][m]) + 1);
	if (~Pre[0][min(n, m)]) ans = max(ans, Solve(Pre[0][min(n, m)], Pre[0][min(n, m)]) + 1);
	return F[P] = ans;
}
int sqz()
{
	int n = read();
	rep(i, 1, n) T[0][i] += (T[1][i] = read());
	rep(i, 1, n) T[0][i] += (T[2][i] = read());
	rep(i, 0, 2)
		rep(j, 1, n) T[i][j] = T[i][j - 1] + T[i][j];
	rep(i, 0, 2)
	{
		Derec.clear(); Derec[0] = 0;
		Pre[i][0] = -1;
		rep(j, 1, n)
		{
			Pre[i][j] = max(Pre[i][j - 1], Derec.count(T[i][j]) ? Derec[T[i][j]] : -1);
			Derec[T[i][j]] = j;
		}
	}
	printf("%d
", Solve(n, n));
}