hdu 3507 斜率优化dp

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#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
;
int q
;
int dp
;
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
);
}
}

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
;
int q
;
int dp

;
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
[m]);
}
}