Bzoj3197/洛谷3296 [SDOI2013]刺客信条assassin(树的重心+树Hash+树形DP+KM)

题面

Bzoj

洛谷

题解

(除了代码均摘自喻队的博客,可是他退役了)

首先固定一棵树,枚举另一棵树,显然另一棵树只有与这棵树同构才有可能产生贡献 如果固定的树以重心为根,那么另一棵树最多就只有重心为根才有可能同构了(可能有两个) 然后就是求改动次数最小值,设$f[x][y]$表示以第一棵树$x$为根的子树内和第二棵树内$y$为根的子树内,达到目标最少需要改动的次数 我们发现只有同构的子树需要决策,我们把同构的子树分别拿出来,我们要做的就是做一个匹配,跑一遍$KM$或者费用流就好了。因为要最小化$f[x][y]$,所以是跑最小完美匹配。 $f[x][y]$要记忆化一下,判断同构用树哈希即可

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using std::sort; using std::vector;
typedef long long ll;

template<typename T>
void read(T &x) {
    int flag = 1; x = 0; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
    while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}

const int N = 1.4e3 + 10, Inf = 1e9 + 7;
void upt0(int &x, int y) { if(x < y) x = y; }
void upt1(int &x, int y) { if(x > y) x = y; }

namespace KM {
	int n, w[N][N], match[N], ret, lx[N], ly[N];
	bool visx[N], visy[N];
	bool Hungary(int x) {
		visx[x] = 1;
		for(int y = 1; y <= n; ++y)
			if(!visy[y] && lx[x] + ly[y] == w[x][y]) {
				visy[y] = true;
				if(!match[y] || Hungary(match[y])) { match[y] = x; return 1; }
			} 
		return 0;
	}
	int main(int opt) {
	    for(int i = 1; i <= n; ++i)
	        for(int j = 1; j <= n; ++j)
	            w[i][j] = opt * w[i][j];
		for(int i = 1; i <= n; ++i) {
			lx[i] = -Inf, ly[i] = 0;
			for(int j = 1; j <= n; ++j) upt0(lx[i], w[i][j]);
		}
		memset(match, 0, sizeof match);
		for(int x = 1; x <= n; ++x)
			while(1) {
				memset(visx, 0, sizeof visx), memset(visy, 0, sizeof visy);
				if(Hungary(x)) break;
				int inc = Inf;
				for(int i = 1; i <= n; ++i)
					if(visx[i])
						for(int j = 1; j <= n; ++j)
							if(!visy[j]) upt1(inc, lx[i] + ly[j] - w[i][j]);
				for(int i = 1; i <= n; ++i) {
					if(visx[i]) lx[i] -= inc;
					if(visy[i]) ly[i] += inc;
				}
			}
		for(int i = 1; i <= n; ++i)
			if(match[i]) ret += w[match[i]][i];
		return opt * ret;
	}
}//KM模板

int n, rt, fir[N], sec[N], f[N][N], c[N][N]; ll hash[N];
int from[N], cnt, to[N << 1], nxt[N << 1];
inline void addEdge(int u, int v) {
	to[++cnt] = v, nxt[cnt] = from[u], from[u] = cnt;
}
int tmp, siz[N];
vector<int> v1[N], v2[N];
inline bool cmp(const int &i, const int &j) { return hash[i] < hash[j]; }

void getrt(int u, int fa) {
	int max_part = 0; siz[u] = 1;
	for(int i = from[u]; i; i = nxt[i]) {
		int v = to[i]; if(v == fa) continue;
		getrt(v, u), siz[u] += siz[v], upt0(max_part, siz[v]);
	} upt0(max_part, n - siz[u]);
	if(max_part < tmp) tmp = max_part, rt = u;
}//寻找树的重心

void dfs(int u, int fa, vector<int> *V) {
	siz[u] = 1, hash[u] = 0, V[u].clear();
	for(int i = from[u]; i; i = nxt[i]) {
		int v = to[i]; if(v == fa) continue;
		dfs(v, u, V), siz[u] += siz[v], V[u].push_back(v);
	} sort(V[u].begin(), V[u].end(), cmp);
	for(int i = V[u].size() - 1; ~i; --i)
		hash[u] = hash[u] * N + hash[V[u][i]];
	hash[u] = hash[u] * N + siz[u];
}//处理各子树hash值以及儿子(将儿子放进一个vector里面)

int dp(int x, int y) {
	if(f[x][y] != -1) return f[x][y];
	f[x][y] = fir[x] ^ sec[y]; int lim = v1[x].size() - 1;
	for(int i = 0; i <= lim; ++i) {
		int j = i;
		while(j < lim && hash[v1[x][j + 1]] == hash[v1[x][i]]) ++j;
		for(int k = i; k <= j; ++k)
			for(int l = i; l <= j; ++l)
				dp(v1[x][k], v2[y][l]);
		for(int k = i; k <= j; ++k)
			for(int l = i; l <= j; ++l)
				KM::w[k - i + 1][l - i + 1] = dp(v1[x][k], v2[y][l]);
        //初始化边权
		KM::ret = 0, KM::n = j - i + 1, f[x][y] += KM::main(-1), i = j;
        //最小化f[x][y]
	} return f[x][y];
}

int main () {
	read(n);
	for(int i = 1, u, v; i < n; ++i)
		read(u), read(v), addEdge(u, v), addEdge(v, u);
	for(int i = 1; i <= n; ++i) read(fir[i]);
	for(int i = 1; i <= n; ++i) read(sec[i]);
	tmp = Inf, getrt(1, 0), dfs(rt, 0, v2); ll tmp = hash[rt];
	int ans = Inf;
	for(int i = 1; i <= n; ++i) {//暴力枚举重心
		dfs(i, 0, v1);
		if(hash[i] == tmp) memset(f, -1, sizeof f), upt1(ans, dp(i, rt));
	} printf("%d\n", ans);
	return 0;
} 

 

posted @ 2019-01-11 13:32  water_mi  阅读(228)  评论(0编辑  收藏