最近在做些樹形DP練練手
原題鏈接
大意就是給你一棵樹，你可以斷開任意數(shù)量的邊，使得剩下的聯(lián)通塊大小乘積最大。

樣例

8
1 2
1 3
2 4
2 5
3 6
3 7
6 8

輸出

18
我首先想的是設(shè)\(f[i]\)表示以\(i\)為根的子樹可獲得的最大收益，但是會發(fā)現(xiàn)這樣無法轉(zhuǎn)移。考慮再加一維，\(f[i][j]\)表示以\(i\)的子樹中，\(i\)所在的聯(lián)通塊大小為\(j\)的最大價值。然后我就傻了，想了半天也沒想起來怎么轉(zhuǎn)移，最后只好看了一眼題解。其實轉(zhuǎn)移好簡單的，貌似是個樹上背包？考慮在\(dfs\)的過程中進(jìn)行\(DP\)，每當(dāng)訪問完一個點\(i\)的子結(jié)點時，累加一下\(sz[i]\)，就枚舉\(j\)，并且用當(dāng)前子結(jié)點的\(DP\)值來更新\(f[i][j]\)。轉(zhuǎn)移方程大概會長成下面這個樣子：

\(f[i][j]=max(f[i][j],f[i][k]*f[v][j-k])\)
（理解的話，就是把之前的大小為\(k\)的聯(lián)通塊和在當(dāng)前子樹中大小為\(j-k\)的聯(lián)通塊拼起來）
同時，我們特別定義\(f[i][0]\)表示以\(i\)為根的子樹可獲得的價值，則他的轉(zhuǎn)移方程比較特殊：
\(f[i][0]=max(f[i][0],f[i][j]*j)\)
如果到這里這道題就結(jié)束的話，代碼會長成下面這樣：

#include <bits/stdc++.h>using namespace std;#define N 700
#define ll long longint n, eid, sz[N+5], head[N+5];
ll f[N+5][N+5];struct Edge {int next, to;
}e[2*N+5];void addEdge(int u, int v) {e[++eid].next = head[u];e[eid].to = v;head[u] = eid;
}void dp(int u, int fa) {sz[u] = 1, f[u][0] = f[u][1] = 1;for(int i = head[u]; i; i = e[i].next) {int v = e[i].to;if(v == fa) continue;dp(v, u);sz[u] += sz[v];for(int j = sz[u]; j >= 1; --j) { //枚舉i所在的聯(lián)通塊大小for(int k = min(j, sz[u]-sz[v]); k >= max(1, j-sz[v]); --k) { //枚舉子樹根結(jié)點所在聯(lián)通塊大小f[u][j] = max(f[u][j], f[u][k]*f[v][j-k]);}}}for(int i = 1; i <= sz[u]; ++i) f[u][0] = max(f[u][0], f[u][i]*i);
}int main() {cin >> n;for(int i = 1, x, y; i <= n-1; ++i) cin >> x >> y, addEdge(x, y), addEdge(y, x);dp(1, 0);cout << f[1][0] << endl;return 0;
}

但是一交上去只有30\(pts\)，一看討論區(qū)，發(fā)現(xiàn)還要用高精度！于是粘了個板子上去，然后就開心的\(MLE\)了￣▽￣。最后把\(int\)換成\(short\)就對了，無語。
粘一下\(AC\)代碼

#include <cstdio>
#include <iostream>
#include <cstring>using namespace std;#define N 700int n, eid;
short sz[N+5], head[N+5];struct Edge {int next, to;
}e[2*N+5];struct bign{ //高精類模板，網(wǎng)上找的static const int maxn = 120;short d[maxn+5];short len;void clean() { while(len > 1 && !d[len-1]) len--; }bign() { memset(d, 0, sizeof(d)); len = 1; }bign(int num) { *this = num; } bign(char* num) { *this = num; }bign operator = (const char* num) {memset(d, 0, sizeof(d)); len = strlen(num);for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';clean();return *this;}bign operator = (int num){char s[20]; sprintf(s, "%d", num);*this = s;return *this;}bign operator + (const bign& b) {bign c = *this; int i;for(i = 0; i < b.len; i++) {c.d[i] += b.d[i];if (c.d[i] > 9) c.d[i] %= 10, c.d[i+1]++;}while (c.d[i] > 9) c.d[i++] %= 10, c.d[i]++;c.len = max(len, b.len);if (c.d[i] && c.len <= i) c.len = i+1;return c;}bign operator - (const bign& b) {bign c = *this; int i;for(i = 0; i < b.len; i++) {c.d[i] -= b.d[i];if (c.d[i] < 0) c.d[i] += 10, c.d[i+1]--;}while (c.d[i] < 0) c.d[i++] += 10, c.d[i]--;c.clean();return c;}bign operator * (const bign& b) const {int i, j; bign c; c.len = len + b.len; for(j = 0; j < b.len; j++)for(i = 0; i < len; i++) c.d[i+j] += d[i]*b.d[j];for(i = 0; i < c.len-1; i++) c.d[i+1] += c.d[i]/10, c.d[i] %= 10;c.clean();return c;}bign operator / (const bign& b) {int i, j;bign c = *this, a = 0;for(i = len - 1; i >= 0; i--) {a = a*10 + d[i];for (j = 0; j < 10; j++) if (a < b*(j+1)) break;c.d[i] = j;a = a - b*j;}c.clean();return c;}bign operator % (const bign& b) {int i, j;bign a = 0;for(i = len - 1; i >= 0; i--) {a = a*10+d[i];for(j = 0; j < 10; j++) if (a < b*(j+1)) break;a = a-b*j;}return a;}bign operator += (const bign& b) {*this = *this+b;return *this;}bool operator <(const bign& b) const {if(len != b.len) return len < b.len;for(int i = len-1; i >= 0; i--)if(d[i] != b.d[i]) return d[i] < b.d[i];return false;}bool operator >(const bign& b) const { return b < *this; }bool operator <= (const bign& b) const { return !(b < *this); }bool operator >= (const bign& b) const { return !(*this < b); }bool operator != (const bign& b) const { return b < *this || *this < b; }bool operator == (const bign& b) const { return !(b < *this) && !(b > *this); }string str() const {char s[maxn] = {};for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';return s;}
}f[N+5][N+5];istream& operator >> (istream& in, bign& x) {string s;in >> s;x = s.c_str();return in;
}ostream& operator << (ostream& out, const bign& x) {out << x.str();return out;
}void addEdge(int u, int v) {e[++eid].next = head[u];e[eid].to = v;head[u] = eid;
}void dp(int u, int fa) {sz[u] = 1, f[u][0] = f[u][1] = 1;for(int i = head[u]; i; i = e[i].next) {int v = e[i].to;if(v == fa) continue;dp(v, u);sz[u] += sz[v];for(int j = sz[u]; j >= 1; --j) {for(int k = min(j, sz[u]-sz[v]); k >= max(1, j-sz[v]); --k) {f[u][j] = max(f[u][j], f[u][k]*f[v][j-k]);}}}for(int i = 1; i <= sz[u]; ++i) f[u][0] = max(f[u][0], f[u][i]*i);
}int main() {cin >> n;for(int i = 1, x, y; i <= n-1; ++i) cin >> x >> y, addEdge(x, y), addEdge(y, x);dp(1, 0);cout << f[1][0] << endl;return 0;
}

轉(zhuǎn)載于:https://www.cnblogs.com/dummyummy/p/9746906.html

總結(jié)

以上是生活随笔為你收集整理的洛谷 P1411 树的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。