HDU-4334 Trouble 哈希表Or有序表查找

这题就是一个数据量大点的a+b+c+d+e = 0的判定，可以用哈希表来做，也可以用来个有序表来寻找解。时间复杂度前者接近O(n^3)，后者则为O(N^3).

代码如下：

哈希表：

#include<iostream>

#include<map>

#define MOD 1000003LL

using namespace std;



typedef long long LL;

const int N = 205;



LL a[6][N];



int head[1000005], idx;



struct Node

{

    LL v;

    int next;    

}e[40005];



void add(LL key)

{

    int rkey, flag = 0;

    if (key > 0) {

        rkey = key % MOD;

    }

    else {

        rkey = (-key) % MOD;

    }

    for (int i = head[rkey]; i != -1; i = e[i].next) {

        if (e[i].v == key) {

            flag  = 1;

            break;    

        }

    }    

    if (!flag) {

        ++idx;

        e[idx].v = key;

        e[idx].next = head[rkey];

        head[rkey] = idx;

    }

}



bool find(LL x)

{

    int rkey;

    if (x > 0) {

        rkey = x % MOD;

    }

    else {

        rkey = (-x) % MOD;

    }

    for (int i = head[rkey]; i != -1; i = e[i].next) {

        if (e[i].v == x) {

            return true;    

        }    

    }    

    return false;

}



int main(){

    int t, n, flag;

    LL key;

    scanf("%d",&t);

    while(t-- && scanf("%d",&n)){

        idx = -1;

        flag = 0;

        memset(head, 0xff, sizeof (head));

        for(int i = 1; i <= 5; i++){

            for(int j = 1; j <= n; j++){

                scanf("%I64d",&a[i][j]);    

            }

        } 

        for(int i = 1; i <= n; i++){

            for(int j = 1; j <= n; j++){

                key = a[1][i] + a[2][j];

                add(key);

            } 

        }

        for(int i = 1; i <= n && !flag; i++){ 

            for(int j = 1; j <= n && !flag; j++){

                for(int k = 1; k <= n; k++){

                    if (find(-(a[3][i]+a[4][j]+a[5][k]))) {

                        flag = 1;

                        break;

                    }

                }    

            }

        }

        puts( flag? "Yes" : "No");

    }

    return 0;

}

有序表查找：将a+b的所有组合放到一个表中，c+d放到另一个表中，都排序去重，再定义两个指针，分别指向第一个表的首，第二个表的尾。再向两边走。

#include <cstdlib>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

typedef long long int Int64;



int N, fidx, sidx;



Int64 seq[5][205];



Int64 flist[400005], slist[400005];



int main()

{

    int T, ca = 0, flag, p1, p2;

    Int64 sum, obj;

    scanf("%d", &T);

    while (T--) {

        fidx = sidx = flag = 0;

        scanf("%d", &N);

        for (int i = 0; i < 5; ++i) {

            for (int j = 1; j <= N; ++j) {

                scanf("%I64d", &seq[i][j]);        

            }

        }

        for (int i = 1; i <= N; ++i) {

            for (int j = 1; j <= N; ++j) {

                flist[++fidx] = seq[0][i] + seq[1][j];

                slist[++sidx] = seq[2][i] + seq[3][j];

            }

        }

        sort(flist+1, flist+1+fidx);

        sort(slist+1, slist+1+sidx);

        fidx = unique(flist+1, flist+1+fidx) - (flist+1);

        sidx = unique(slist+1, slist+1+sidx) - (slist+1);

        for (int i = 1; i <= N && !flag; ++i) {

            p1 = 1, p2 = sidx;

            obj = -seq[4][i];

            while (p1 <= fidx && p2 >= 1) {

                sum = flist[p1] + slist[p2];

                if (obj > sum) {

                    ++p1;

                }    

                else if (obj < sum) {

                    --p2;

                }

                else {

                    flag = 1;

                    break;

                }

            }

        }

        printf(flag ? "Yes\n" : "No\n");

    }

    return 0;    

}