hdu 4334 Trouble

Trouble

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3854    Accepted Submission(s): 1222Problem DescriptionHassan is in trouble. His mathematics teacher has given him a very difficult problem called 5-sum. Please help him.
The 5-sum problem is defined as follows: Given 5 sets S_1,...,S_5 of n integer numbers each, is there a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0? 

InputFirst line of input contains a single integer N (1≤N≤50). N test-cases follow. First line of each test-case contains a single integer n (1<=n<=200). 5 lines follow each containing n integer numbers in range [-10^15, 1 0^15]. I-th
line denotes set S_i for 1<=i<=5. 

OutputFor each test-case output "Yes" (without quotes) if there are a_1 in S_1,...,a_5 in S_5 such that a_1+...+a_5=0, otherwise output "No". 

Sample Input

221 -11 -11 -11 -11 -131 2 3-1 -2 -34 5 6-1 3 2-4 -10 -1

 

Sample Output

NoYes

 

Source2012 Multi-University Training Contest 4
 思路：合并a[0]、a[1]，a[2]、a[3]，得到三個數組a[4]、s1[ ]、s2[ ]，對a[4]枚舉，然後線上性時間內算出是否存在i、j,使得s1[i]+s2[j]+a[k]=0。線性演算法如下：將s1[ ]、s2[ ]排序，用兩個變數p1，p2分別表示訪問到s1[ ]、s2[ ]的位置，p1從前往後，p2從後往前。如果a[k]+s1[p1]+s2[p2]>0，則p2往後移，因為此時可以得出p2與其他的p1匹配都只會使三個數的和變大，所以p2沒有必要與其他的p1匹配後算了，同理可得a[k]+s1[p1]+s2[p2]<0的情況。
舉一反三：這個演算法可以解決這種類型的問題：給你三個數組，從三個數組中任選一個數，判斷這三個數的和是否能為一個值。
代碼：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#define maxn 205#define maxx 40005using namespace std;int n,m,ans,cxx0,cxx1,cxx2;long long a[5][maxn];long long s1[maxx],s2[maxx];long long xx[maxx],yy[maxx],zz[maxn];void solve()      // 合并a[0]、a[1],a[2]、a[3] 將相同的去掉 剪枝{    int i,j,cnt;    for(i=0;i<5;i++)    {        sort(a[i],a[i]+n);    }    cnt=-1;    for(i=0;i<n;i++)    {        for(j=0;j<n;j++)        {            cnt++;            s1[cnt]=a[0][i]+a[1][j];            s2[cnt]=a[2][i]+a[3][j];        }    }    zz[0]=a[4][0];    cxx0=0;    for(i=1;i<n;i++)    {        if(a[4][i]!=a[4][i-1])        {            cxx0++;            zz[cxx0]=a[4][i];        }    }    sort(s1,s1+cnt+1);    xx[0]=s1[0];    cxx1=0;    for(i=1;i<=cnt;i++)    {        if(s1[i]!=s1[i-1])        {            cxx1++;            xx[cxx1]=s1[i];        }    }    sort(s2,s2+cnt+1);    yy[0]=s2[0];    cxx2=0;    for(i=1;i<=cnt;i++)    {        if(s2[i]!=s2[i-1])        {            cxx2++;            yy[cxx2]=s2[i];        }    }}bool isok(){    int i,j,k;    long long nz;    for(k=0;k<=cxx0;k++)   // 對a[4]枚舉    {        nz=zz[k];        for(i=0,j=cxx2;i<=cxx1||j>=0;)  // 線性時間判斷是否存在a+b+c=0        {            if(nz+xx[i]+yy[j]>0)            {                if(j>0) j--;                else break ;            }            else if(nz+xx[i]+yy[j]<0)            {                if(i<cxx1) i++;                else break ;            }            else return true ;        }    }    return false ;}int main(){    int i,j,t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0;i<5;i++)        {            for(j=0;j<n;j++)            {                scanf("%I64d",&a[i][j]);            }        }        solve();        if(isok()) printf("Yes\n");        else printf("No\n");    }    return 0;}