Codeforces 321E Ciel and Gondolas

传送门：http://codeforces.com/problemset/problem/321/E

【题解】

首先有一个$O(n^2k)$的dp。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10, N = 4000 + 5;
const int mod = 1e9+7;
const ll inf = 1e17;

inline int getint() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = (x<<3) + (x<<1) + ch - ‘0‘, ch = getchar();
    return x;
}

int n, K, a[N][N], d[N][N];
ll f[805][N];

int main() {
    n = getint(), K = getint();
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=n; ++j)
            a[i][j] = a[i-1][j] + a[i][j-1] - a[i-1][j-1] + getint();
    for (int i=1; i<=n; ++i) {
        d[i][i] = 0;
        for (int j=i+1; j<=n; ++j)
            d[i][j] = (a[j][j] - a[j][i-1] - a[i-1][j] + a[i-1][i-1])/2;
    }
    for (int i=1; i<=n; ++i) f[0][i] = inf;
    for (int i=1; i<=K; ++i) {
        for (int j=1; j<=n; ++j) {
            f[i][j] = inf;
            for (int k=0; k<j; ++k)
                f[i][j] = min(f[i][j], f[i-1][k] + d[k+1][j]);
        }
    }
    cout << f[K][n];

    return 0;
}

然后发现每层当n向右移动的时候，决策点一定向右移动

（可能可以观察+证明）

然后用整体二分trick就可以了。复杂度$O(Knlogn)$。

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10, N = 4000 + 5;
const int mod = 1e9+7;
const ll inf = 1e17;

inline int getint() {
    int x = 0; char ch = getchar();
    while(!isdigit(ch)) ch = getchar();
    while(isdigit(ch)) x = (x<<3) + (x<<1) + ch - ‘0‘, ch = getchar();
    return x;
}

int n, K, a[N][N], d[N][N];
ll f[805][N], *F, *G;

inline void solve(int l, int r, int al, int ar) {
    if(l > r) return ;
    int mid = l+r>>1; ll mx = inf, t; int pos = 0;
    for (int j=al; j<=ar && j<mid; ++j)
        if((t = G[j] + d[j+1][mid]) < mx) mx = t, pos = j;
    F[mid] = mx;
    solve(l, mid-1, al, pos);
    solve(mid+1, r, pos, ar);
}

int main() {
    n = getint(), K = getint();
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=n; ++j)
            a[i][j] = a[i-1][j] + a[i][j-1] - a[i-1][j-1] + getint();
    for (int i=1; i<=n; ++i) {
        d[i][i] = 0;
        for (int j=i+1; j<=n; ++j)
            d[i][j] = (a[j][j] - a[j][i-1] - a[i-1][j] + a[i-1][i-1])/2;
    }
    for (int i=1; i<=n; ++i) f[0][i] = inf;
    for (int i=1; i<=K; ++i) {
        F = f[i], G = f[i-1];
        solve(1, n, 0, n);
    }
    cout << f[K][n];

    return 0;
}