HDU 2829 Lawrence（DP+四边形不等式优化）

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题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=2829

思路：大神地址：http://www.tuicool.com/articles/nMzAzi6

四边形不等式：http://wenku.baidu.com/view/be418243a8956bec0975e3bf.html

待写代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long LL;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;LL dp[1010][1010],cost[1010][1010];int a[1010];int s[1010][1010];int main(){    int n,m;    while(scanf("%d%d",&n,&m))    {        if(m + n == 0)            break;        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        /*memset(cost,0,sizeof(cost));        for(int i=1; i<=n; i++)        {            for(int j=i+1; j<=n; j++)            {                cost[i][j] = cost[i][j-1];                for(int k=i; k<=j; k++)                {                    cost[i][j] += a[j] * a[k];                }            }        }*/        for(int i=1; i<=n; i++)        {            LL sum = 0;            cost[i][i] = 0;            for(int j=i+1; j<=n; j++)            {                sum += a[j-1];                cost[i][j] = cost[i][j-1] + sum * a[j];            }        }        for(int i=1; i<=n; i++)        {            dp[0][i] = cost[1][i];            s[0][i] = 0;            s[i][n+1] = n;        }        for(int i=1; i<=m; i++)        {            for(int j=n; j>=1; j--)//这层循环是倒着来着，和01背包一个道理            {                dp[i][j] = cost[i][j];                for(int k=s[i-1][j]; k<=s[i][j+1]; k++)                {                    dp[i][j] = min(dp[i][j],dp[i-1][k]+cost[k+1][j]);                    s[i][j] = k;                }            }        }        printf("%I64d\n",dp[m][n]);    }    return 0;}

超时代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long LL;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1using namespace std;LL dp[1010][1010],cost[1010][1010];int a[1010];int main(){    int n,m;    while(scanf("%d%d",&n,&m))    {        if(m + n == 0)            break;        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        /*memset(cost,0,sizeof(cost));        for(int i=1; i<=n; i++)        {            for(int j=i+1; j<=n; j++)            {                cost[i][j] = cost[i][j-1];                for(int k=i; k<=j; k++)                {                    cost[i][j] += a[j] * a[k];                }            }        }*/        for(int i=1; i<=n; i++)        {            LL sum = 0;            cost[i][i] = 0;            for(int j=i+1; j<=n; j++)            {                sum += a[j-1];                cost[i][j] = cost[i][j-1] + sum * a[j];            }        }        for(int i=1; i<=n; i++)        {            dp[0][i] = cost[1][i];        }        for(int i=1; i<=m; i++)        {            for(int j=n; j>=1; j--)//这层循环是倒着来着，和01背包一个道理            {                dp[i][j] = cost[i][j];                for(int k=i; k<j; k++)                {                    dp[i][j] = min(dp[i][j],dp[i-1][k]+cost[k+1][j]);                }            }        }        printf("%I64d\n",dp[m][n]);    }    return 0;}

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