Luogu1941 飞扬的小鸟
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这题的 dp 还是比较显然的
听说是个完全背包,大概转移是差不多的
就从当前层顺着枚举 j 往大去更新同层的就好了
其实这样每次往高处转移的就是下面的前缀最小值
值得注意的是题意要模拟的是游戏
所以显然不能先掉下去在在同一步中往上飞
所以转移顺序是不能乱的
就是先转移往上飞的在转移往下掉的
好像多开一维也行,不过没有必要了
代码:
#include <algorithm> #include <iostream> #include <cstring> #include <cstdlib> #include <cctype> #include <cstdio> #include <cmath> using namespace std; const int MAX_N = 10005, MAX_M = 1005; int n, m, k, max_dst; int x[MAX_N], y[MAX_N], pos[MAX_N], pre_sum[MAX_N]; int bot[MAX_N], top[MAX_N], f[MAX_N][MAX_M]; bool has_stk[MAX_N]; int main() { scanf("%d%d%d", &n, &m, &k); for (int i = 1; i <= n; ++i) scanf("%d%d", &x[i], &y[i]); for (int i = 1; i <= k; ++i) { scanf("%d", &pos[i]); scanf("%d%d", &bot[pos[i]], &top[pos[i]]); has_stk[pos[i]] = true; } for (int i = 1; i < n; ++i) pre_sum[i] = pre_sum[i - 1] + has_stk[i]; memset(f, 0x3f, sizeof(f)); for (int i = 1; i <= m; ++i) f[0][i] = 0; for (int i = 1; i <= n; ++i) { for (int j = x[i] + 1; j <= m; ++j) { max_dst = (f[i][j] < 0x3f3f3f3f) ? i : max_dst; f[i][j] = min(f[i][j], min(f[i - 1][j - x[i]], f[i][j - x[i]]) + 1); } for (int j = m - x[i]; j <= m; ++j) { max_dst = (f[i][j] < 0x3f3f3f3f) ? i : max_dst; f[i][m] = min(f[i][m], min(f[i - 1][j], f[i][j]) + 1); } for (int j = 1; j <= m - y[i]; ++j) { max_dst = (f[i][j] < 0x3f3f3f3f) ? i : max_dst; f[i][j] = min(f[i][j], f[i - 1][j + y[i]]); } if (has_stk[i]) { for (int j = 1; j <= bot[i]; ++j) f[i][j] = 0x3f3f3f3f; for (int j = top[i]; j <= m; ++j) f[i][j] = 0x3f3f3f3f; } } register int min_res = 0x3f3f3f3f; if (!has_stk[n]) { for (int i = 1; i <= m; ++i) if (f[n][i] < 0x3f3f3f3f) { max_dst = n; min_res = min(min_res, f[n][i]); } } else { for (int i = bot[n] + 1; i < top[n]; ++i) if (f[n][i] < 0x3f3f3f3f) { max_dst = n; min_res = min(min_res, f[n][i]); } } if (max_dst == n) { puts("1"); printf("%d", min_res); } else { printf("0 %d", pre_sum[max_dst] - has_stk[max_dst]); } return 0; }
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