标签：set linker pre int tree cos [] swa string
题意：给定上一棵树和一个排列，然后问你把这个排列分成m个连续的部分，每个部分的大小的是两两相邻的LCA的最小深度，问你最小是多少。
析：首先这个肯定是DP，然后每个部分其实就是里面最小的那个LCA的深度。很容易知道某个区间的值肯定是 [li, li+1] .. [ri-1, ri]这些区间之间的一个，并且我们还可以知道，举个例子，1 2 3 4 5 6 如果知道分成两部分 其中 2 和 6 是最优的，那么中间的 3 4 5 ，这三个数其实属于哪个区间都无所谓，所以对于第 i 个数只有三种可能。
dp[i[j] 表示前 i 个数分成 j 个区间
第一种：它自己属于单独的区间，dp[i][j] = min{ dp[i-1][j-1] + deep[a[i]] }
第二种：它和前面那个数属于一个区间，dp[i][j] = min{ dp[i-2][j-1] + deep[lca(a[i], a[i-1])] }
第三种：它对任何区间都没有贡献，那么无所谓了 dp[i][j] = min{ dp[i-1][j] }
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
//#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 3e5 + 20;
const int maxm = 100 + 10;
const ULL mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int to, next;
};
Edge edge[maxn<<1];
int head[maxn], cnt;
void addEdge(int u, int v){
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u] = cnt++;
}
int a[maxn];
int dp[3][maxn];
int dep[maxn], p[20][maxn];
void dfs(int u, int fa, int d){
dep[u] = d;
p[0][u] = fa;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to;
if(v == fa) continue;
dfs(v, u, d + 1);
}
}
void init(){
ms(p, -1);
dfs(1, -1, 1);
FOR(k, 0, 19) for(int v = 1; v <= n; ++v){
if(p[k][v] < 0) p[k+1][v] = -1;
else p[k+1][v] = p[k][p[k][v]];
}
}
int LCA(int u, int v){
if(dep[u] > dep[v]) swap(u, v);
for(int k = 0; k < 20; ++k)
if(dep[v] - dep[u] >> k & 1) v = p[k][v];
if(u == v) return u;
for(int k = 19; k >= 0; --k)
if(p[k][u] != p[k][v]){
u = p[k][u];
v = p[k][v];
}
return p[0][u];
}
int lca[maxn];
int main(){
while(scanf("%d %d", &n, &m) == 2){
for(int i = 1; i <= n; ++i) scanf("%d", a + i);
ms(head, -1); cnt = 0;
for(int i = 1; i < n; ++i){
int u, v;
scanf("%d %d", &u, &v);
addEdge(u, v);
addEdge(v, u);
}
init(); ms(dp, INF); lca[1] = dep[a[1]];
for(int i = 2; i <= n; ++i) lca[i] = dep[LCA(a[i], a[i-1])];
dp[0][0] = dp[1][0] = dp[2][0] = 0;
for(int i = 1; i <= n; ++i){
int t = min(i, m);
for(int j = 1; j <= t; ++j){
dp[i%3][j] = min(dp[(i-1)%3][j-1] + dep[a[i]], dp[(i-1)%3][j]);
if(i > 1) dp[i%3][j] = min(dp[i%3][j], dp[(i-2)%3][j-1] + lca[i]);
}
}
printf("%d\n", dp[n%3][m]);
}
return 0;
}
HDU 6065 RXD, tree and sequence (LCA+DP)
标签：set linker pre int tree cos [] swa string
原文地址：http://www.cnblogs.com/dwtfukgv/p/7613142.html
