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。
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
注意要记忆化
#include #include #include #include #include #include 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 Derec; map, 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 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\n", Solve(n, n)); }
转载于:https://www.cnblogs.com/WizardCowboy/p/7756262.html
