2017ICPC北京区域赛 Pangu and Stones 区间dp

给$n$个数和一个上下限$L$和$R$，每次可以将一段连续的$k$个数合并，代价为数值之和，且要满足$k\in [L,R]$。求最终合并为1堆的最小代价。

与合并石子类似，定义$dp[i][j][k]$为区间$[i,j]$剩下$k$堆的最小代价，因为只有$l\le k\le r$才能合并，所以将$k=1$单独转移：
dp[i][j][k]=min⁡{dp[i][p][1]+dp[p+1][j][k−1]}      k≥2dp[i][j][1]=min⁡{dp[i][j][k]+s[i]−s[j−1]}      l≤k≤r\begin{aligned} &dp[i][j][k] = \min\{ dp[i][p][1] + dp[p + 1][j][k - 1]\} & \ \ \ \ \ \ k\geq2\\ & dp[i][j][1] =\min\{ dp[i][j][k] + s[i]-s[j-1]\} & \ \ \ \ \ \ l\leq k \leq r\\ \end{aligned}​dp[i][j][k]=min{dp[i][p][1]+dp[p+1][j][k−1]}dp[i][j][1]=min{dp[i][j][k]+s[i]−s[j−1]}​      k≥2      l≤k≤r​

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<cstdlib>

#define inf 0x3f3f3f3f
#define ll long long
#define mem(a, x) memset(a,x,sizeof(a))
typedef std::pair<ll, ll> Pll;
const int N = 110;
using namespace std;

ll gcd(ll p, ll q) { return q == 0 ? p : gcd(q, p % q); }

ll dp[N][N][N], a[N], s[N];
int n, l, r;

inline void init() {
    for (int i = 1; i <= n; i++) cin >> a[i], s[i] = s[i - 1] + a[i];
    mem(dp, inf);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    while (cin >> n >> l >> r) {
        init();
        for (int d = 1; d <= n; d++)
            for (int i = 1, j; j = i + d - 1, j <= n; i++) {
                dp[i][j][d] = 0;
                for (int k = 2; k <= d; k++) {
                    for (int p = i; p <= j; p++)
                        dp[i][j][k] = min(dp[i][j][k], dp[i][p][1] + dp[p + 1][j][k - 1]);
                    if (l <= k && k <= r)
                        dp[i][j][1] = min(dp[i][j][1], dp[i][j][k] + s[j] - s[i - 1]);
                }
            }
        printf("%lld\n", dp[1][n][1] >= inf ? 0 : dp[1][n][1]);
    }
    return 0;

}