POJ 3671 Dining Cows 简单DP

http://poj.org/problem?id=3671

简单动态规划

opt[i][0]第i个数后1的个数；opt[i][1] 第i个数前2的个数

状态方程

opt[i][0]=opt[i-1][0]-1 //value[i]=1 opt[i][0]=opt[i-1][0] //value[i]=2 opt[i][1]=opt[i-1][1] //value[i-1]=1 opt[i][1]=opt[i-1][1]+1 //value[i-1]=2

 

Dining Cows

Time Limit: 1000MS		Memory Limit: 65536K

Description

The cows are so very silly about their dinner partners. They have organized themselves into two groups (conveniently numbered 1 and 2) that insist upon dining together in order, with group 1 at the beginning of the line and group 2 at the end. The trouble starts when they line up at the barn to enter the feeding area.

Each cow i carries with her a small card upon which is engraved D_i (1 ≤ D_i ≤ 2) indicating her dining group membership. The entire set of N (1 ≤ N ≤ 30,000) cows has lined up for dinner but it's easy for anyone to see that they are not grouped by their dinner-partner cards.

FJ's job is not so difficult. He just walks down the line of cows changing their dinner partner assignment by marking out the old number and writing in a new one. By doing so, he creates groups of cows like 112222 or 111122 where the cows' dining groups are sorted in ascending order by their dinner cards. Rarely he might change cards so that only one group of cows is left (e.g., 1111 or 222).

FJ is just as lazy as the next fellow. He's curious: what is the absolute minimum number of cards he must change to create a proper grouping of dining partners? He must only change card numbers and must not rearrange the cows standing in line.

Input

* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 describes cow i's dining preference with a single integer: D_i

Output

* Line 1: A single integer that is the minimum number of cards Farmer John must change to assign the cows to eating groups as described.

Sample Input

7
2
1
1
1
2
2
1

Sample Output

2

47ms代码

/* Author : yan
 * Question : POJ 3671 Dining Cows
 * Date && Time : Thursday, January 20 2011 01:13 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
int value[30001];
int opt[30001][2];
///opt[i][0]第i个数后1的个数；opt[i][1] 第i个数前2的个数
int main()
{
	freopen("input","r",stdin);
	int n;
	int cnt;
	int ans;
	memset(opt,0,sizeof(opt));
	scanf("%d",&n);
	int i;
	for(i=0;i<n;i++)
	{
		scanf("%d",&value[i]);
		if(value[i]==1) cnt++;
	}

	if(value[0]==1) opt[0][0]=cnt-1;
	else opt[0][0]=cnt;
	opt[0][1]=0;

	for(i=1;i<n;i++)
	{
		if(value[i]==1) opt[i][0]=opt[i-1][0]-1;
		else opt[i][0]=opt[i-1][0];
		if(value[i-1]==1) opt[i][1]=opt[i-1][1];
		else opt[i][1]=opt[i-1][1]+1;
	}
	ans=0xFFFFFFF;
	for(i=0;i<n;i++)
	{
		if(ans>opt[i][0]+opt[i][1]) ans=opt[i][0]+opt[i][1];
	}
	printf("%d",ans);
	return 0;
}
 

稍微修改优化后代码 16ms

/* Author : yan
 * Question : POJ 3671 Dining Cows
 * Date && Time : Thursday, January 20 2011 01:13 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
int value[30001];
int opt[30001][2];
///opt[i][0]第i个数后1的个数；opt[i][1] 第i个数前2的个数
int main()
{
	freopen("input","r",stdin);
	int n;
	int cnt;
	int ans;
	memset(opt,0,sizeof(opt));
	scanf("%d",&n);
	int i;
	for(i=0;i<n;i++)
	{
		scanf("%d",&value[i]);
		if(value[i]==1) cnt++;
	}

	if(value[0]==1) opt[0][0]=cnt-1;
	else opt[0][0]=cnt;
	opt[0][1]=0;

	for(i=1;i<n;i++)
	{
		opt[i][0]=opt[i-1][0];
		if(value[i]==1) opt[i][0]--;
		opt[i][1]=opt[i-1][1];
		if(value[i-1]==2) opt[i][1]++;
	}
	ans=0xFFFFFFF;
	for(i=0;i<n;i++)
	{
		if(ans>opt[i][0]+opt[i][1]) ans=opt[i][0]+opt[i][1];
	}
	printf("%d",ans);
	return 0;
}