经典的环形Dp问题,定义dp[i][j]为将第i堆到第j堆石子全部合并后所用的最小花费。则有如下状态转移方程:
dp[i][j]=min(dp[i][j],dp[i][k−1]+dp[k][j]+w[i][j])(k∈(i,j])
通过四边形不等式优化以后,可转化为:
dp[i][j]=min(dp[i][j],dp[i][k−1]+dp[k][j]+w[i][j])(k∈[s[i][j−1],s[i+1][j]])
其中s[i][j]=max(k|dp[i][j]=dp[i][k−1]+dp[k][j]+w[i][j])
代码如下:
/*
ID: Sunshine_cfbsl
LANG: C++
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 105, INF = 1000000000;
int ans1, ans2, n, w[MAXN*2][MAXN*2], dp[MAXN*2][MAXN*2], s[MAXN*2][MAXN*2], f[MAXN*2][MAXN*2];
int main() {
int i, d, k;
scanf("%d", &n);
for(i = 1; i <= n; i++) {
scanf("%d", &w[i][i]);
w[i+n][i+n] = w[i][i];
dp[i+n][i+n] = dp[i][i] = f[i+n][i+n] = f[i][i] = 0;
s[i+n][i+n] = i+n;
s[i][i] = i;
}
for(d = 1; d < n; d++) {
for(i = 1; i + d < 2*n; i++) {
w[i][i+d] = w[i+1][i+d]+w[i][i];
dp[i][i+d] = INF;
f[i][i+d] = -INF;
k = s[i][i+d-1];
if(k == i) k++;
for(; k <= s[i+1][i+d]; k++) {
if(dp[i][i+d] >= dp[i][k-1]+dp[k][i+d]+w[i][i+d]) {
dp[i][i+d] = dp[i][k-1]+dp[k][i+d]+w[i][i+d];
s[i][i+d] = k;
}
}
f[i][i+d] = max(f[i+1][i+d], f[i][i+d-1]) + w[i][i+d];
}
}
ans1 = INF;
ans2 = -INF;
for(i = 1; i+n-1 < 2*n; i++) {
ans1 = min(ans1, dp[i][i+n-1]);
ans2 = max(ans2, f[i][i+n-1]);
}
printf("%d\n%d\n", ans1, ans2);
return 0;
}