hdu 3507 斜率优化dp

http://www.cnblogs.com/kuangbin/archive/2012/08/26/2657650.html

#include<math.h>
#include<string.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll __int64
#define N 500050
int n,m;
int sum[N];
int q[N];
int dp[N];
int getup(int j,int k)
{
	return dp[j]+sum[j]*sum[j]-(dp[k]+sum[k]*sum[k]);
}
int getdown(int j,int k)
{
	return 2*(sum[j]-sum[k]);
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i,j,k;
		int l=0,r=0;
		sum[0]=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&k);
			sum[i]=k+sum[i-1];
		}
		q[r++]=0;
		dp[0]=0;
		for(i=1;i<=n;i++)
		{
			while(l+1<r && getup(q[l+1],q[l])<=sum[i]*getdown(q[l+1],q[l]))
				l++;
			j=q[l];
			dp[i]=dp[j]+(sum[i]-sum[j])*(sum[i]-sum[j])+m;
			while(l+1<r && getup(i,q[r-1])*getdown(q[r-1],q[r-2])<=getup(q[r-1],q[r-2])*getdown(i,q[r-1]))
				r--;
			q[r++]=i;
		}
		printf("%d\n",dp[n]);
	}
}

hdu 2829

//二位优化
题意：有一段铁路有n个站，每个站可以往其他站运送粮草，现在要炸掉m条路使得粮草补给最小，粮草补给的公式是将每个站能收到的粮草的总和。
4----5-----1-----2
Its Strategic Value is 4*5 + 4*1 + 4*2 + 5*1 + 5*2 + 1*2 = 49.
4----5       1-----2
The Strategic Value of the remaining railroad is 4*5 + 1*2 = 22. 
4      5-----1------2
The Strategic Value of the remaining railroad is 5*1 + 5*2 + 1*2 = 17.
dp[i][j]表示到i为止的前面炸了j条路，的最大消耗，，用总消耗减去最大消耗等于最小花费。
dp[i][j]=max(dp[i][k-1]+sum[i]*(sum[i]-sum[j]));

#include<math.h>
#include<string.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll __int64
#define N 1010
int n,m;
int sum[N];
int q[N];
int dp[N][N];
int getup(int j,int x,int k)
{
	return sum[j]*sum[j]-dp[j][x-1]-(sum[k]*sum[k]-dp[k][x-1]);
}
int getdown(int j,int k)
{
	return (sum[j]-sum[k]);
}
int main()
{
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==0 &&m==0)
			break;
		int i,j,k;
		int l=0,r=0;
		sum[0]=0;
		int ans=0;
		for(i=1;i<=n;i++)
		{
			scanf("%d",&k);
			sum[i]=k+sum[i-1];
			ans+=k*sum[i-1];
		}
		memset(dp,0,sizeof(dp));
		int x;
		for(x=1;x<=m;x++)
		{
			l=r=0;
			q[r++]=x;
			for(i=x+1;i<=n;i++)
			{
				while(l+1<r && getup(q[l+1],x,q[l])<=sum[i]*getdown(q[l+1],q[l]))
					l++;
				j=q[l];
				dp[i][x]=dp[j][x-1]+sum[j]*(sum[i]-sum[j]);
				while(l+1<r && getup(i,x,q[r-1])*getdown(q[r-1],q[r-2])<=getup(q[r-1],x,q[r-2])*getdown(i,q[r-1]))
					r--;
				q[r++]=i;
			}
		}
		printf("%d\n",ans-dp[n][m]);
	}
}