poj 1717 Dominoes 背包

Dominoes

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6464 Accepted: 2131

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

背包问题，数组的应用。

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>using namespace std;typedef long long ll;const int INF =0x3f3f3f3f;const int maxn=1000    ;int n,V,st;int val[maxn+5];int  dp[maxn+5][12010];int fun(int x){    return x+6000;}int ans,best;void work(){    int x,y;    V=st=0;    for(int i=1;i<=n;i++)    {        scanf("%d%d",&x,&y);        val[i]=x-y;        st+=val[i];        V+=abs(val[i]);    }    memset(dp[0],0x3f,sizeof dp[0]);    dp[0][fun(st)]=0;    for(int i=1;i<=n;i++)    {        int tval=-2*val[i];        for(int v=-V;v<=V;v++)        {            dp[i][fun(v)]=dp[i-1][fun(v)];            if( -V<=v-tval&&v-tval<=V)            dp[i][fun(v)]=min(dp[i][fun(v)],dp[i-1][fun(v-tval)]+1);        }    }    int best=INF,ans=INF;    for(int v=-V;v<=V;v++) if(dp[n][fun(v)]!=INF)    {        if( abs(v)<best )        {            best=abs(v);            ans=dp[n][fun(v)];        }        else if(abs(v)==best)        {            ans=min(ans,dp[n][fun(v)]);        }    }    cout<<ans<<endl;}int main(){    while(~scanf("%d",&n) )    work();    return 0;}