设f[i]表示i个鼹鼠出现后,打死鼹鼠的最大值
动态转移方程:f[i]=max{f[j]+1},
条件:abs(x[i]-x[j])+abs(y[i]-y[j])<=time[i]-time[j] (j<i)
代码
#include <cstdio> #include <iostream> #define N 1001 #define M 10001 #define abs(x) ((x) < 0 ? -(x) : (x)) #define max(x, y) ((x) > (y) ? (x) : (y)) int n, m, ans; int s[M][3], f[N], mx[N]; inline int read() { int x = 0, f = 1; char ch = getchar(); for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1; for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0'; return x * f; } int main() { int i, j, k, x, y, z; n = read(); m = read(); for(k = 1; k <= m; k++) { s[k][0] = read(); s[k][1] = read(); s[k][2] = read(); f[k] = 1; for(i = k - 1; i >= 1; i--) { if(f[k] >= mx[i] + 1) break; if(s[k][0] - s[i][0] >= abs(s[k][1] - s[i][1]) + abs(s[k][2] - s[i][2])) f[k] = max(f[k], f[i] + 1); } ans = max(ans, f[k]); mx[k] = max(mx[k - 1], f[k]); } printf("%d ", ans); return 0; }