以hdu3480为例学会斜率优化&&四边形优化

来源：互联网发布：结构体包含数组memset 编辑：程序博客网时间：2024/06/05 07:42

1、斜率优化，要通过状态转移方程算出现行变化的y1 y2 x1 x2 然后斜率比较，要维护成凸的形状
2、第一次接触的就是四边形优化，不过理解不深，其实四边形优化就是记录上一次最有值转移过来的是哪个值，然后下次转移时就直接从这个位置开始，对时间有很大的优化。

3、两者比较起来后者比较好写些，第一种方法需要一丁点数学功底（没差），如果四边形熟悉的话，用四边形会更容易写出来。

斜率优化：

/** this code is made by LinMeiChen* Problem:* Type of Problem:* Thinking:* Feeling:*/#include<iostream>#include<algorithm>#include<stdlib.h>#include<string.h>#include<stdio.h>#include<math.h>#include<string>#include<vector>#include<queue>#include<list>using namespace std;typedef long long lld;typedef unsigned int ud;#define oo 0x3f3f3f3f#define maxn 10010#define maxm 5010int a[maxn];int dp[maxm][maxn];int q[maxn], front, rear;int main(){    int n, m, T;    scanf("%d", &T);    for (int cas = 1; cas <= T; cas++)    {        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++)            scanf("%d", &a[i]);        sort(a + 1, a + 1 + n);        for (int i = 1; i <= n; i++)            dp[1][i] = (a[i] - a[1])*(a[i] - a[1]);        for (int i = 2; i <= m; i++)        {            front = rear = 0;            q[rear++] = i - 1;            for (int j = i; j <= n; j++)            {                while (front + 1 < rear)                {                    int k1 = q[front];                    int k2 = q[front + 1];                    int x1 = a[k1 + 1];                    int x2 = a[k2 + 1];                    int y1 = dp[i - 1][k1] + x1*x1;                    int y2 = dp[i - 1][k2] + x2*x2;                    if (y2 - y1 <= 2 * a[j] * (x2 - x1))                        front++;                    else                        break;                }                int k = q[front];                dp[i][j] = dp[i - 1][k] + (a[j] - a[k + 1])*(a[j] - a[k + 1]);                while (front + 1 < rear)                {                    int k1 = q[rear - 2];                    int k2 = q[rear - 1];                    int k3 = j;                    int x1 = a[k1 + 1];                    int x2 = a[k2 + 1];                    int x3 = a[k3 + 1];                    int y1 = dp[i - 1][k1] + x1*x1;                    int y2 = dp[i - 1][k2] + x2*x2;                    int y3 = dp[i - 1][k3] + x3*x3;                    if ((y3 - y2)*(x2 - x1) <= (y2 - y1)*(x3 - x2))                        rear--;                    else                        break;                }                q[rear++] = j;            }        }        printf("Case %d: %d\n", cas, dp[m][n]);    }    return 0;}

四边形优化：

/** this code is made by LinMeiChen* Problem:* Type of Problem:* Thinking:* Feeling:*/#include<iostream>#include<algorithm>#include<stdlib.h>#include<string.h>#include<stdio.h>#include<math.h>#include<string>#include<vector>#include<queue>#include<list>using namespace std;typedef long long lld;typedef unsigned int ud;#define oo 0x3f3f3f3f#define maxn 10010#define maxm 5010int a[maxn];int dp[maxm][maxn];int mark[maxm][maxn];int main(){    int n, m, T;    scanf("%d", &T);    for (int cas = 1; cas <= T; cas++)    {        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++)            scanf("%d", &a[i]);        sort(a + 1, a + 1 + n);        for (int i = 1; i <= n; i++)        {            dp[1][i] = (a[i] - a[1])*(a[i] - a[1]);            mark[1][i] = 1;        }        for (int i = 2; i <= m; i++)        {            mark[i][n + 1] = n - 1;            for (int j = n; j >= i; j--)            {                dp[i][j] = -1;                for (int k = mark[i - 1][j]; k <= mark[i][j + 1]; k++)                {                    int temp = (a[j] - a[k + 1])*(a[j] - a[k + 1]);                    if (dp[i][j] == -1 || dp[i - 1][k] + temp < dp[i][j])                    {                        dp[i][j] = dp[i - 1][k] + temp;                        mark[i][j] = k;                    }                }            }        }        printf("Case %d: %d\n", cas, dp[m][n]);    }    return 0;}

0 0