poj 2785 双向搜索+哈希

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000KTotal Submissions: 16271 Accepted: 4718Case Time Limit: 5000MS

Description

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

Input

The first line of the input file contains the size of the lists n (this value can be as large as 4000). We then have n lines containing four integer 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 -4626 -38 -10 62-32 -54 -6 45

Sample Output

5

Hint

Sample Explanation: 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

先枚举出所有a[i]+b[j]的值，存到哈希表中。

再枚举所有tmp=c[i]+d[j]的值，检查-tmp是否在哈希表中，返回个数。

3200+ms.如果用排序加二分查找5000+ms。。。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;#define maxn 4001#define M 21331131const int inf =(1<<29);int a[maxn], b[maxn],c[maxn],d[maxn];int n;struct node{    node(){}    node(int v, int n, int  c){val = v; next =n; cnt = c;}    int val;    int next;    int cnt;};node h[maxn*maxn];int first[M], top;      //邻接表写法，first为表头指针void insert(int v){    int pos = (v+inf)%M;    int tmp = -1;    for(int i = first[pos]; i!=-1; i=h[i].next)    if(h[i].val == v){        h[i].cnt++;        return;    }    h[top] = node(v, first[pos], 1);    first[pos] = top++;}int find(int v){    int pos = (v+inf)%M;    for(int i = first[pos]; i != -1; i=h[i].next)        if(v == h[i].val)            return h[i].cnt;    return 0;}int main(){    while(scanf("%d", &n)!=EOF){        memset(first, -1, sizeof(first));        top = 0;        for(int i = 0; i < n; i++)            scanf("%d%d%d%d", a+i, b+i, c+i, d+i);        for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++)            insert(a[i]+b[j]);        long long ans = 0;        int tmp;        for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++){            tmp = c[i]+d[j];            ans += find(-tmp);        }        printf("%lld\n", ans);    }    return 0;}