Codeforces 868F

题意：给定一个序列要求分成k个连续子序列，使得子序列的cost之和最小，子序列的cost定义为子序列中数对的两个数相同的数对个数, 如cost(1,1) = 1, oost(1,1,1,3) = cost(1,1,1) = 3*2/2 = 1+2 = 3; 

题解：对于朴素的dp[i][j]表示1~j分为i个块的最小cost，有O(k*n^2)的算法dp[i][j] = min{dp[i-1][k]+cost(k+1,j), k < j}, 设给定j最优的k为P(j),不难发现P(j)单调,于是分治,solve(l,r,L,R)表示已知对于 l <= j <= r, 有 L <= p(j) <= R. 令m = (l+r)/2, 每次通过枚举计算出P(m)以及dp[i][m], 再分治计算solve(l,m-1,L,p(m))和solve(m+1,r,p(m),R).

再考虑cost(i,j), 不难发现cost(i,j)可以O(1)转移到相邻区间, 且在分治过程中每次solve区间移动O(n)次，所以莫队之即可，总复杂度O(knlogn)

#include <iostream>#include <cstdio>#include <cctype>#include <algorithm>#include <cstring>#include <string>#include <cmath>#include <vector>#include <set>#include <stack>#include <sstream>#include <queue>#include <map>#include <functional>#include <bitset>using namespace std;#define pb push_back#define mk make_pair#define ll long long#define ull unsigned long long#define pii pair<int, int>#define mk make_pair#define fi first#define se second#define ALL(A) A.begin(), A.end()#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)#define repr(i, n) for(int (i)=(int)(n);(i)>=0;(i)--)#define repab(i,a,b) for(int (i)=(int)(a);(i)<=(int)(b);(i)++)#define reprab(i,a,b) for(int (i)=(int)(a);(i)>=(int)(b);(i)--)#define m ((l+r)/2)#define sc(x) scanf("%d", &x)#define pr(x) printf("x:%d\n", x)#define fastio ios::sync_with_stdio(0), cin.tie(0)#define frein freopen("in.txt", "r", stdin)#define freout freopen("out.txt", "w", stdout)#define freout1 freopen("out1.txt", "w", stdout)#define lb puts("")#define PI M_PI#define debug cout<<"???"<<endl#define mid ((l+r)>>1)const ll mod = 1000000007;//const int INF = 0x3f3f3f3f;const ll INF = 0x3f3f3f3f3f3f3f3f;const double eps = 1e-6;template<class T> T gcd(T a, T b){if(!b)return a;return gcd(b,a%b);}const int maxn = 1e5+10;int n,k,a[maxn],cnt[maxn],le = 1, ri = 0, now, nxt;ll sum, dp[2][maxn];void query(int l, int r){    while(ri < r){        sum += 1LL*cnt[a[++ri]];        cnt[a[ri]]++;        //printf("ri:%d, sum:%I64d\n", ri, sum);    }    while(ri > r){        cnt[a[ri]]--;        sum -= 1LL*cnt[a[ri--]];    }    while(le > l){        sum += 1LL*cnt[a[--le]];        cnt[a[le]]++;    }    while(le < l){        cnt[a[le]]--;        sum -= 1LL*cnt[a[le++]];    }    //printf("query(%d,%d):%I64d\n", l, r, sum);}void solve(int l, int r, int L, int R){    if(l > r) return;    dp[nxt][m] = INF;    int M, ed = min(R,m);    for(int i = L; i <= ed; i++){        query(i, m);        if(dp[now][i-1] + sum < dp[nxt][m]){            dp[nxt][m] = dp[now][i-1] + sum;            M = i;        }    }    //printf("solve(%d,%d,%d,%d), dp[%d]:%I64d\n", l,r,L,R,m,dp[nxt][m]);    solve(l,m-1,L,M);    solve(m+1,r,M,R);}int main(){    //frein;    while(cin >> n >> k){        for(int i = 1; i <= n; i++) sc(a[i]);        memset(cnt, 0, sizeof(cnt));        now = 0; nxt = 1;        le = 1; ri = 0; sum = 0;        memset(dp, INF, sizeof(dp));        dp[now][0] = 0;        for(int i = 0; i < k; i++){            solve(1,n,1,n);            swap(now,nxt);        }        printf("%I64d\n", dp[now][n]);    }    return 0;}