POJ-2785 （hash，二分）

POJ 2785

4 Values whose Sum is 0

Time Limit: 15000MS Memory Limit: 228000K
Total Submissions: 26941 Accepted: 8122
Case 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 -46
26 -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).

思路

普通的dfs不行了，这时候可以对四个数组，两两结合进行查找。

• 二分
• Hash

二分

二分STl忘了？

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#define N 4004
using namespace std;
int a[N], b[N], c[N], d[N];
int l[N * N + 10], r[N * N + 10];
int main() {
//  freopen("in.txt", "r", stdin);
    int n;
    while (scanf("%d", &n) != EOF) {
        for (int i = 0; i < n; i++) {
            scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
        }
        //对四个数组两两处理 
        int now = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int t = a[i] + b[j];
                l[now++] = t;
            }
        }
        sort(l, l + now);

        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                int t = c[i] + d[j];
                if(-t > l[now - 1]) continue;
                int ll = 0, rr = now - 1;
                // 二分找第一个大于等于key
                //   int *first = lower_bound(l, l + now, -t);
                while (ll < rr) {
                    int mid = (ll + rr) / 2;
                    if(l[mid] < -t) {
                        ll = mid + 1;
                    }else if(l[mid] >= -t) {
                        rr = mid;
                    }
                }
                int first = ll;     
                ll = 0, rr = now - 1;
                // 二分找第一个大于key 
                //  int *last = upper_bound(l, l + now, -t);
                while (ll < rr) {
                    int mid = (ll + rr) / 2;
                    if(l[mid] <= -t) {
                        ll = mid + 1;
                    }else if(l[mid] > -t) {
                        rr = mid;
                    }
                }
                if(l[ll] == -t)     ll++;
                int last = ll;

                ans += last - first;    
            }
        }       

        cout << ans << endl;
    }   
    return 0;
}

Hash

不知道什么是Hash？

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<cmath>
#include<string.h>
#define N 4004
using namespace std;
int a[N], b[N], c[N], d[N];
struct ac{
    static const int mask = 0x7fffff;
    int sum[N * N], vis[N * N];
    void clear(){
        memset(sum, 0, sizeof(sum));
    } 
    int& operator [](int x){
        int i;
        for(i = x & mask; sum[i] && vis[i] != x; i = (i + 1) & mask);
        vis[i] = x;
        return sum[i];
    }
}Hash;

int main() {
    int n;
    while (scanf("%d", &n) != EOF) {

        for (int i = 0; i < n; i++) {
            scanf("%d%d%d%d", &a[i], &b[i], &c[i], &d[i]);
        }
        Hash.clear();
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                Hash[a[i] + b[j]]++; 
            }
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                ans += Hash[-c[i] - d[j]];
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}