HDU 2829 斜率优化DP Lawrence

题意：n个数之间放m个障碍，分隔成m+1段。对于每段两两数相乘再求和，然后把这m+1个值加起来，让这个值最小。

设：

d(i, j)表示前i个数之间放j个炸弹能得到的最小值

sum(i)为前缀和，cost(i)为前i个数两两相乘之和。

则有状态转移方程：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 1000 + 10;
const int INF = 0x3f3f3f3f;

int n, m;
int a[maxn], sum[maxn], cost[maxn];
int d[maxn][maxn], s[maxn][maxn];

int inline w(int k, int j)
{
return cost[j] - cost[k-1] - sum[k-1] * (sum[j] - sum[k-1]);
}

int main()
{
while(scanf("%d%d", &n, &m) == 2 && n)
{
m++;

for(int i = 1; i <= n; i++) scanf("%d", a + i);
for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i];
for(int i = 2; i <= n; i++) cost[i] = cost[i-1] + sum[i-1] *  a[i];

memset(d, 0, sizeof(d));
for(int i = 1; i <= m; i++)
for(int j = i + 1; j <= n; j++) d[i][j] = INF;

for(int i = 1; i <= n; i++) { d[1][i] = cost[i]; s[1][i] = 0; }
for(int i = 2; i <= m; i++)
{
s[i][n+1] = n;
for(int j = n; j >= i; j--)
{
for(int k = s[i-1][j]; k <= s[i][j+1]; k++)
{
int tmp = d[i-1][k] + w(k+1, j);
if(tmp < d[i][j])
{ d[i][j] = tmp; s[i][j] = k; }
}
}
}

printf("%d\n", d[m]
);
}

return 0;
}

代码君