PKU 2785 4 Values whose Sum is 0

4 Values whose Sum is 0

Time Limit: 15000MS
Memory Limit: 228000KTotal Submissions: 6014
Accepted: 1456Case Time Limit: 5000MS

Description

TheSUM problem can be formulated as follows: given four lists A, B, C, Dof integer values, compute how many quadruplet (a, b, c, d ) ∈ A x B xC x D are such that a + b + c + d = 0 . In the following, we assumethat all lists have the same size n .

Input

Thefirst line of the input file contains the size of the lists n (thisvalue can be as large as 4000). We then have n lines containing fourinteger values (with absolute value as large as 2²⁸ ) that belong respectively to A, B, C and D .

Output

For each input file, your program has to write the number quadruplets whose sum is zero.

Sample Input

6
-45 22 42 -16
-41 -27 56 30
-36 53 -37 77
-36 30 -75 -46
26 -38 -10 62
-32 -54 -6 45

Sample Output

5

Hint

SampleExplanation: Indeed, the sum of the five following quadruplets is zero:(-45, -27, 42, 30), (26, 30, -10, -46), (-32, 22, 56, -46),(-32, 30,-75, 77), (-32, -54, 56, 30).

Source

Southwestern Europe 2005

偶用2分过了以后又用HASH过了次，发现HASH就是快- -!
2分5000MS+，HASH 3000MS+
差距啊。。。

2分就是先把1，2两列的和先算出来，然后再把3，4两列的和也算出来，最后二分查找得到一个解，注意解的附近也可能有解存在！

HASH的时候开静态数会很快~偶用的是最土的取MOD完接表。

下面是2分代码：

1. //代码不长不过效率相对来说不怎么高...,复杂度2*N^2 *(logN + 1)
2. #include<iostream>
3. #include<algorithm>
4. using namespace std;
5. #define N 4000
6. #define M 16000000
7. int a[N],b[N],c[N],d[N],ab[M],cd[M];
8. int main()
9. {
10.     int n,i,j,k,l,r,mid,cnt,ablen,cdlen;
11.     while(scanf("%d",&n)!=EOF)
12.     {
13.     cnt=ablen=cdlen=0;
14.     for(i=0;i<n;i++)
15.     {
16.         scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
17.     }
18.     for(i=0;i<n;i++)
19.         for(j=0;j<n;j++)
20.             ab[ablen++]=a[i]+b[j];
21.     for(i=0;i<n;i++)
22.         for(j=0;j<n;j++)
23.             cd[cdlen++]=-c[i]-d[j];
24.     sort(cd,cd+cdlen);
25.     for(i=0;i<ablen;i++)
26.     {
27.         l=0;
28.         r=cdlen-1;
29.         while(l<=r)
30.         {
31.             mid=(l+r)/2;
32.             if(cd[mid]==ab[i])
33.             {
34.                 cnt++;
35.                 for(k=mid+1;k<cdlen;k++)
36.                     if(cd[k]!=ab[i])break;
37.                     else cnt++;
38.                 for(k=mid-1;k>=0;k--)
39.                     if(cd[k]!=ab[i])break;
40.                     else cnt++;
41.                 break;
42.             }
43.             if(cd[mid]<ab[i])
44.                 l=mid+1;
45.             else r=mid-1;
46.         }
47.     }
48.     printf("%d/n",cnt);
49.     }
50.     return 0;
51. }

之后的HASH写的就比较顺利，有了ZX大牛的指点,1次AC，效率比他还高- -!!

1. #include<iostream>
2. using namespace std;
3. #define N 4000
4. #define M 17100000
5. const int p=1048576;//(1<<20) 为了取MOD的时候效率高点~
6. struct Node
7. {
8.     int val,cnt,next;
9. };
10. Node hash[M];
11. bool flag[M];
12. int a[N][4],hlen=p,cnt=0;
13. inline int gn()
14. {
15.     char buf=getchar();
16.     int sgn=(buf=='-'?-1:1),ret=0;
17.     if(buf!='-')ungetc(buf,stdin);
18.     while((buf=getchar())!=' '&&buf!='/n')
19.         ret=ret*10+(buf-48);
20.     return ret*sgn;
21. }
22. void ins(int val,int key)
23. {
24.     if(!flag[key])
25.     {
26.         flag[key]=true;
27.         hash[key].cnt=1;
28.         hash[key].val=val;
29.         hash[key].next=-1;
30.         return;
31.     }
32.     while(key!=-1)
33.     {
34.         if(val==hash[key].val)
35.         {
36.             hash[key].cnt++;
37.             return;
38.         }
39.         if(hash[key].next==-1)break;
40.         key=hash[key].next;
41.     }
42.     hash[key].next=++hlen;
43.     hash[hlen].val=val;
44.     hash[hlen].cnt=1;
45.     hash[hlen].next=-1;
46. }
47. void fnd(int val,int key)
48. {
49.     if(!flag[key])
50.         return;
51.     while(key!=-1)
52.     {
53.         if(val==hash[key].val)
54.         {
55.             cnt+=hash[key].cnt;
56.             return;
57.         }
58.         key=hash[key].next;
59.     }
60. }
61. int main()
62. {
63.     int n,i,j,tmp;
64.     bool nag;
65.     n=gn();
66.     for(i=0;i<n;i++)
67.         for(j=0;j<4;j++)
68.             a[i][j]=gn();
69.     for(i=0;i<n;i++)
70.         for(j=0;j<n;j++)
71.         {
72.             tmp=a[i][0]+a[j][1];
73.             nag=tmp<0;
74.             if(nag)tmp*=-1;
75.             ins(a[i][0]+a[j][1],tmp&(p-1));
76.         }
77.     for(i=0;i<n;i++)
78.         for(j=0;j<n;j++)
79.         {
80.             tmp=-a[i][2]-a[j][3];
81.             nag=tmp<0;
82.             if(nag)tmp*=-1;
83.             fnd(-a[i][2]-a[j][3],tmp&(p-1));
84.         }
85.     printf("%d/n",cnt);
86.     return 0;
87. }