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：

 

设0 ≤ l < k < i，且k比l更优，有不等式：

 

整理得到，注意不等号方向：

 最后变成了斜率的形式，下面就用一个队列维护即可。

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

const int maxn = 1000 + 10;

int n, m;

int sum[maxn], cost[maxn];
int d[maxn][maxn];

int head, tail;
int Q[maxn];

int j;

int inline Y(int i)
{
    return d[i][j-1] + sum[i] * sum[i] - cost[i];
}

int inline DY(int x1, int x2) { return Y(x2) - Y(x1); }

int inline DX(int x1, int x2) { return sum[x2] - sum[x1]; }

int inline DP(int x1, int x2) { return d[x1][j-1] + cost[x2] - cost[x1] - sum[x1]*(sum[x2]-sum[x1]); }

int main()
{
    while(scanf("%d%d", &n, &m) == 2 && n)
    {
        for(int i = 1; i <= n; i++) scanf("%d", sum + i);
        for(int i = 2; i <= n; i++) sum[i] += sum[i - 1];
        for(int i = 2; i <= n; i++) cost[i] = cost[i-1] + (sum[i]-sum[i-1])*sum[i-1];

        for(int i = 1; i <= n; i++) d[i][0] = cost[i];
        for(j = 1; j <= m; j++)
        {
            head = tail = 0;
            Q[tail++] = 0;
            for(int i = 1; i <= n; i++)
            {
                while(head + 1 < tail && DY(Q[head], Q[head+1]) <= sum[i] * DX(Q[head], Q[head+1])) head++;
                d[i][j] = DP(Q[head], i);
                while(head + 1 < tail && DY(Q[tail-1], i) * DX(Q[tail-2], Q[tail-1]) <= DY(Q[tail-2], Q[tail-1]) * DX(Q[tail-1], i)) tail--;
                Q[tail++] = i;
            }
        }

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

    return 0;
}

 

贴一个四边形不等式优化的代码对比一下：

#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][n]);
    }

    return 0;
}