POJ 1717 Dominoes

Description

A domino is a flat, thumbsized tile, the face of which is divided into two squares, each left blank or bearing from one to six dots. There is a row of dominoes laid out on a table: 

The number of dots in the top line is 6+1+1+1=9 and the number of dots in the bottom line is 1+5+3+2=11. The gap between the top line and the bottom line is 2. The gap is the absolute value of difference between two sums. 

Each domino can be turned by 180 degrees keeping its face always upwards. 

What is the smallest number of turns needed to minimise the gap between the top line and the bottom line? 

For the figure above it is sufficient to turn the last domino in the row in order to decrease the gap to 0. In this case the answer is 1. 
Write a program that: computes the smallest number of turns needed to minimise the gap between the top line and the bottom line.

Input

The first line of the input contains an integer n, 1 <= n <= 1000. This is the number of dominoes laid out on the table. 

Each of the next n lines contains two integers a, b separated by a single space, 0 <= a, b <= 6. The integers a and b written in the line i + 1 of the input file, 1 <= i <= 1000, are the numbers of dots on the i-th domino in the row, respectively, in the top line and in the bottom one. 

Output

Output the smallest number of turns needed to minimise the gap between the top line and the bottom line.

Sample Input

46 11 51 31 2

Sample Output

1

Source

CEOI 1997

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01背包~

翻车了……

记录一个maxx值表示上下最大差值。

用f[i][j]表示差值为j-maxx时的最小步数，如果f[i][j]==n表示不能得到这个差值。其中i这一维滚动。

用d(i)来表示i+maxx，结果最后求值的时候忘记了，写成了d(maxx+i)……WA到怀疑人生……所以一定要细心呀！

#include<cstdio>#include<cstring>#include<iostream>using namespace std;#define d(u) (u+maxx)int n,a[1001],maxx,f[2][20001],inf;bool kkz;int read(){int x=0,f=1;char ch=getchar();while(ch<'0' || ch>'9') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0' && ch<='9') {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f; }int main(){n=read();for(int i=1;i<=n;i++){a[i]=read()-read();maxx+=a[i]<0 ? -a[i]:a[i];}for(int i=0;i<=2*maxx;i++) f[kkz][i]=n+1;inf=n+1;f[kkz][d(a[1])]=0;f[kkz][d(-a[1])]=1;for(int i=2;i<=n;i++){kkz^=1;for(int j=0;j<=2*maxx;j++) f[kkz][j]=n+1;for(int j=0;j<=2*maxx;j++)  if(f[kkz^1][j]!=inf)  {  f[kkz][j+a[i]]=min(f[kkz][j+a[i]],f[kkz^1][j]);f[kkz][j-a[i]]=min(f[kkz][j-a[i]],f[kkz^1][j]+1);  }}for(int i=0;;i++)  if(f[kkz][d(i)]!=inf || f[kkz][d(-i)]!=inf)  {  printf("%d\n",min(f[kkz][d(i)],f[kkz][d(-i)]));break;  }return 0;}