题目大意：给你N个数，要求你分成M个部分，每部分的得分为，假设该部分为i, j, k, 那么的分就为i * j + i * k + j * k
问划分后，至少能得多少分
解题思路：设dp[i][j]为前面i个数，分成j个部分能得到得最小得分
得转移方程dp[i][j] = dp[k][j - 1] + cost[i] - cost[k] - (sum[i] - sum[k]) * sum[k]
这里的cost[i]表示的是前i个数为1个部分的情况下所能得到的分
因为cost[i] - cost[k]后，还多了sum[k] * (sum[i] - sum[k])，所以后面要减去
现在设l > k，且在l比j更优
则dp[k][j - 1] + cost[i] - cost[k] - sum[k] * (sum[i] - sum[k]) >= dp[l][j - 1] + cost[i] - cost[l] - sum[l] * (sum[i] - sum[l])
化简可得
sum[i] >= (dp[l][j - 1] - cost[l] + sum[l] ^ 2 - dp[k][j - 1] + cost[k] - sum[k] ^ 2) / (sum[l] - sum[k])
这样就得到斜率优化的式子了
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int sum[N], cost[N], que[N], dp[N][N];
int getUp(int j, int k, int boom) {
return dp[j][boom] - cost[j] + sum[j] * sum[j] - (dp[k][boom] - cost[k] + sum[k] * sum[k]);
}
int getDown(int j, int k) {
return sum[j] - sum[k];
}
void solve() {
for (int i = 1; i <= n; i++) {
dp[i][0] = cost[i];
dp[i][i - 1] = 0;
}
int head, tail;
for (int boom = 1; boom <= m; boom++) {
head = tail = 0;
que[tail++] = boom;
for (int i = boom + 1; i <= n; i++) {
while (head + 1 < tail && getUp(que[head + 1], que[head], boom - 1) <= getDown(que[head + 1], que[head]) * sum[i])
head++;
int t = que[head];
dp[i][boom] = dp[t][boom - 1] + cost[i] - cost[t] + sum[t] * sum[t] - sum[i] * sum[t];
while (head + 1 < tail && getUp(i, que[tail - 1], boom - 1) * getDown(que[tail - 1], que[tail - 2]) <= getUp(que[tail - 1], que[tail - 2], boom - 1) * getDown(i, que[tail - 1]))
tail--;
que[tail++] = i;
}
}
printf("%d\n", dp[n][m]);
}
void init() {
sum[0] = cost[0] = 0;
for (int i = 1; i <= n; i++) {
scanf("%d", &sum[i]);
cost[i] = cost[i - 1] + sum[i - 1] * sum[i];
sum[i] += sum[i - 1];
}
}
int main() {
while (scanf("%d%d", &n, &m) != EOF && n + m) {
init();
solve();
}
return 0;
}
