道路
题目背景:
分析:记忆化搜索
没什么难度的裸DP,直接定义dp[i][j][k]表示以i为根的子树中,i到跟有j条没有修过的公路,k条没有修过的铁路,然后如果是乡村则直接返回计算过的答案,否则
dp[i][j][k] = std::min(dp[lc[i]][j][k] + dp[rc[i]][j][k + 1]
, dp[lc[i]][j+ 1][k], dp[rc[i]][j][k])
(lc[i]表示i的公路下的儿子,rc[i]是铁路下的)
直接记忆化搜索就好了,注意要开long long。
时间复杂度O(n * 40 * 40)
Source:
/* created by scarlyw */ #include <cstdio> #include <string> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <cctype> #include <vector> #include <set> #include <queue> #include <ctime> #include <bitset> inline char read() { static const int IN_LEN = 1024 * 1024; static char buf[IN_LEN], *s, *t; if (s == t) { t = (s = buf) + fread(buf, 1, IN_LEN, stdin); if (s == t) return -1; } return *s++; } ///* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = read(), iosig = false; !isdigit(c); c = read()) { if (c == -1) return ; if (c == '-') iosig = true; } for (x = 0; isdigit(c); c = read()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int OUT_LEN = 1024 * 1024; char obuf[OUT_LEN]; char *oh = obuf; inline void write_char(char c) { if (oh == obuf + OUT_LEN) fwrite(obuf, 1, OUT_LEN, stdout), oh = obuf; *oh++ = c; } template<class T> inline void W(T x) { static int buf[30], cnt; if (x == 0) write_char('0'); else { if (x < 0) write_char('-'), x = -x; for (cnt = 0; x; x /= 10) buf[++cnt] = x % 10 + 48; while (cnt) write_char(buf[cnt--]); } } inline void flush() { fwrite(obuf, 1, oh - obuf, stdout), oh = obuf; } /* template<class T> inline void R(T &x) { static char c; static bool iosig; for (c = getchar(), iosig = false; !isdigit(c); c = getchar()) if (c == '-') iosig = true; for (x = 0; isdigit(c); c = getchar()) x = ((x << 2) + x << 1) + (c ^ '0'); if (iosig) x = -x; } //*/ const int MAXN = 20000 + 10; const int MAXX = 40 + 2; int n; long long dp[MAXN][MAXX][MAXX]; int a[MAXN], b[MAXN], c[MAXN], lc[MAXN], rc[MAXN]; inline long long calc(int id, int x, int y) { return (long long)c[id] * (a[id] + x) * (b[id] + y); } inline long long dfs(int cur, int x, int y) { if (cur < 0) return calc(-cur, x, y); if (dp[cur][x][y] != -1) return dp[cur][x][y]; dp[cur][x][y] = std::min(dfs(lc[cur], x, y) + dfs(rc[cur], x, y + 1) , dfs(lc[cur], x + 1, y) + dfs(rc[cur], x, y)); return dp[cur][x][y]; } inline void read_in() { R(n), memset(dp, -1, sizeof(dp)); for (int i = 1; i < n; ++i) R(lc[i]), R(rc[i]); for (int i = 1; i <= n; ++i) R(a[i]), R(b[i]), R(c[i]); } int main() { freopen("road.in", "r", stdin); freopen("road.out", "w", stdout); read_in(); std::cout << dfs(1, 0, 0); return 0; }