解題思路做完Hdu的2829，然後再看這題，壹切變得如此簡單，用兩種方法解。
狀態轉移方程為: dp[i] = min(dp[j] + (sum[i]-sum[j])^2 + m) (j < i);
方法壹：dp[i] = dp[j] + (sum[i]-sum[j])^2 + m = dp[j] + sum[i] * sum[i] + sum[j] * sum[j] - 2 * sum[i] * sum[j] + m;
設y = dp[j] + sum[j] * sum[j],x = sum[j]，那麽原式等於:dp[i] = y + 2 * sum[i] * x + m + sum[i] * sum[i],然後套下斜率優化DP模板即可ac。
方法二:方法二使用的優化技巧類似於四邊形不等式，用個s[i] 記錄dp[i]由前面的哪個狀態轉移過來，然後枚舉的時候只要枚舉s[i-1] 到i-1就可以了。
第二種方法似乎比第壹種要慢壹些，常數比較大。
AC源代碼
#include <stdio.h>
#include <string.h>
#define MAX 510000
#define int64 long long
struct point{
int64 x,y,c;
}pot[MAX];
int n,m,arr[MAX];
int64 sum[MAX],dp[MAX];
int qu[MAX],head,tail;
int CheckIt(int x,int y,int z) {
point p0 = pot[x],p1 = pot[y],p2 = pot[z];
return (p0.x -p1.x) * (p0.y - p2.y) - (p0.y - p1.y) * (p0.x - p2.x) <= 0;
}
int NotBest(int x,int y,int k) {
point p0 = pot[x],p1 = pot[y];
return p0.y - k * p0.x > p1.y - k * p1.x;
}
int64 Solve_DP() {
head = tail = 0;
qu[tail] = 0;
pot[0].x = pot[0].y = 0;
for (int i = 1; i <= n; ++i) {
pot[i].x = sum[i-1];
pot[i].y = dp[i-1] + sum[i-1] * sum[i-1];
while (head <= tail - 1 &&
CheckIt(qu[tail-1],qu[tail],i)) tail--;
qu[++tail] = i;
while (head + 1 <= tail &&
NotBest(qu[head],qu[head+1],2 * sum[i])) head++;
int k = qu[head];
dp[i] = pot[k].y - 2 * sum[i] * pot[k].x + sum[i] * sum[i] + m;
}
return dp[n];
}
int64 Solve_DP2() {
for (int64 mmin,i = 1; i <= n; ++i) {
mmin = -1;
for (int j = qu[i-1]; j < i; ++j)
if (mmin == -1 ||
dp[j] + (sum[i] - sum[j]) * (sum[i] - sum[j]) < mmin) {
mmin = dp[j] + (sum[i] - sum[j]) * (sum[i] - sum[j]);
qu[i] = j;
}
dp[i] = mmin + m;
}
return dp[n];
}
int main()
{
int i,j,k;
while (scanf("%d%d",&n,&m) != EOF) {
for (i = 1; i <= n; ++i)
scanf("%d",&arr[i]),sum[i] = arr[i] + sum[i-1];
int64 ans = Solve_DP2();
printf("%lld\n",ans);
}
}