文章目錄

• • 1.二分查找法
  • • 1.1 思路：
    • 1.2 AC代碼
  • 2.hash查找法
  • • 2.1 思路：
    • 2.2 Wrong Answer 代碼
    • 2.3 Time Limit Exceeded 代碼
    • 2.4 偷懶失敗，hash_map在poj中不存在
    • 2.5 哈希表+二叉查找樹（超時）
    • 2.6 AC代碼（哈希+數(shù)組法）

題目鏈接： http://poj.org/problem?id=2785

題目大意：給定不超過4000行4個數(shù)，每列挑出來1個，使之和為0，有多少種方案。

1.二分查找法

1.1 思路：

• 對左邊兩列的所有和求出來，右邊兩列所有的和的求出來再取負(fù)
• 右側(cè)兩列的值排序（進(jìn)行二分查找）
• 對左邊所有的和在右邊的和中進(jìn)行二分查找，并查找前后都滿足要求的，計數(shù)即得答案
  

1.2 AC代碼

/*** @description: 求四個數(shù)相加等于0的組合種數(shù)* @author: michael ming* @date: 2019/5/8 22:18* @modified by:*/ #include <iostream> #include <memory.h> #include <algorithm>using namespace std; int a[4001], b[4001], c[4001], d[4001]; int ab[4000*4000+1], cd[4000*4000+1]; //存儲a+b，c+d int findSameValue(int value, int maxindex)//二分查找 {int low = 0, high = maxindex, mid, index, count=0;while(low <= high){mid = low + (high - low)/2;if(value == cd[mid]) //查找前后滿足要求的個數(shù){index = mid - 1;while(index >= 0 && value == cd[index--])count++;index = mid + 1;while(index <= high && value == cd[index++])count++;return count + 1;}else if(value < cd[mid])high = mid - 1;elselow = mid + 1;}return 0; } int main() {int line, k=0;cin >> line;memset(ab, 0, sizeof(ab));memset(cd, 0, sizeof(cd));for(int i = 0; i < line; ++i){cin >> a[i] >> b[i] >> c[i] >> d[i];}for(int i = 0; i < line; ++i){for(int j = 0; j < line; ++j){ab[k] = a[i]+b[j];cd[k++] = -(c[i]+d[j]);}}sort(cd,cd+k); //二分查找的必要條件，有序int result = 0;for(int i = 0; i <= k-1; ++i)result += findSameValue(ab[i], k-1);cout << result << endl;return 0; }

2.hash查找法

2.1 思路：

• 對左邊兩列的所有和求出來，右邊兩列所有的和的求出來再取負(fù)
• 對左側(cè)兩列的和存入哈希表
• 對右邊兩列的和在哈希表中查找

2.2 Wrong Answer 代碼

(未解決沖突，并且hash后的值可能為負(fù)，作為數(shù)組下標(biāo)不行，所以錯誤)

/*** @description: 4個數(shù)和為0的方案數(shù)，哈希法* @author: michael ming* @date: 2019/5/9 22:30* @modified by: */ #include <iostream> #include <memory.h> using namespace std; int a[4001], b[4001], c[4001], d[4001]; int ab[4000*4000+1], cd[4000*4000+1]; //存儲a+b，c+d int *hasht = new int[16000057]; int offset = 1000000000; int hashfunc(int &value) {int mod = 16000057;return (value%mod + value/mod)%mod; } int main() {int line, k=0, value;cin >> line;memset(hasht, 0, sizeof(hasht));for(int i = 0; i < line; ++i){cin >> a[i] >> b[i] >> c[i] >> d[i];}for(int i = 0; i < line; ++i){for(int j = 0; j < line; ++j){ab[k] = a[i]+b[j];value = ab[k]+offset;hasht[hashfunc(value)]++;cd[k++] = -(c[i]+d[j]);}}int result = 0;for(int i = 0; i < k; ++i){value = cd[i]+offset;if(hasht[hashfunc(value)])result += hasht[hashfunc(value)];}cout << result << endl;delete [] hasht;return 0; }

2.3 Time Limit Exceeded 代碼

/*** @description: 4個數(shù)和為0的方案數(shù)，哈希法* @author: michael ming* @date: 2019/5/9 22:30* @modified by: */ #include <iostream> #include <math.h> using namespace std; struct linkedNode //鏈表節(jié)點(diǎn) {pair<int, int> data;linkedNode *next;linkedNode():next(NULL), data(make_pair(0,0)){} }; class linkedList //鏈表 { public:linkedNode *head;linkedList(){head = new linkedNode(); //表頭哨兵}~linkedList(){delete head;} }; class linkedHash { private:linkedList *htList; //散列表鏈表數(shù)組int bucket; //散列表桶個數(shù) public:linkedHash(int m):bucket(m){htList = new linkedList [bucket] ();}~linkedHash(){for(int i = 0; i < bucket; ++i){linkedNode *p = htList[i].head->next, *q = p;while(q != NULL){p = q;q = q->next;delete p;}}delete [] htList;}int hash(const int &key) const{return abs(key%bucket); //留余數(shù)法}linkedNode* find(const int &x) const{int i = hash(x);linkedNode *p = htList[i].head->next, *q = htList[i].head;while(p && p->data.first != x){q = p;p = p->next;}return q; //返回找到元素的前一個節(jié)點(diǎn)，或者沒有找到,返回最后一個元素}linkedNode* insert(const int &x){int i = hash(x);linkedNode *p = htList[i].head, *q = p;while(q != NULL){p = q;q = q->next;if(q && q->data.first == x){q->data.second++;return q;}}p->next = new linkedNode();p->next->data.first = x;p->next->data.second++;return p->next;} }; int a[4001], b[4001], c[4001], d[4001]; int ab[4000*4000+1], cd[4000*4000+1]; //存儲a+b，c+d int main() {linkedHash ht(16000057);int line, k=0;cin >> line;for(int i = 0; i < line; ++i){cin >> a[i] >> b[i] >> c[i] >> d[i];}for(int i = 0; i < line; ++i){for(int j = 0; j < line; ++j){ab[k] = a[i]+b[j];ht.insert(ab[k]);cd[k++] = -(c[i]+d[j]);}}int result = 0;linkedNode* p;for(int i = 0; i < k; ++i){p = ht.find(cd[i])->next;if(p && p->data.first == cd[i])result += p->data.second;}cout << result << endl;return 0; }

2.4 偷懶失敗，hash_map在poj中不存在

/*** @description: 4個數(shù)和為0的方案數(shù)，哈希法* @author: michael ming* @date: 2019/5/9 22:30* @modified by: */ #include <iostream> #include <hash_map> //#include <unordered_map> using namespace std; int a[4001], b[4001], c[4001], d[4001]; int ab[4000*4000+1], cd[4000*4000+1]; //存儲a+b，c+d int main() {__gnu_cxx::hash_map<int, int> ht; // std::unordered_map<int, int> ht;int line, k=0;cin >> line;for(int i = 0; i < line; ++i){cin >> a[i] >> b[i] >> c[i] >> d[i];}for(int i = 0; i < line; ++i){for(int j = 0; j < line; ++j){ab[k] = a[i]+b[j];ht[ab[k]]++;cd[k++] = -(c[i]+d[j]);}}int result = 0;for(int i = 0; i < k; ++i){result += ht[cd[i]];}cout << result << endl;return 0; }

2.5 哈希表+二叉查找樹（超時）

可能是建立二叉樹插入節(jié)點(diǎn)new太耗時了。

/*** @description: 4個數(shù)和為0的方案數(shù)，哈希法* @author: michael ming* @date: 2019/5/9 22:30* @modified by:*/ #include <iostream> #include <math.h> using namespace std; class BSTNode { public:int data, count;BSTNode *left, *right;BSTNode():count(1), left(NULL), right(NULL){}BSTNode(const int& d, BSTNode *l = NULL, BSTNode *r = NULL){data = d; count = 1; left = l; right = r;} }; class BST { private:BSTNode* root;int nodeLen; public:BST():root(NULL){}~BST(){}void clear(BSTNode* nodeP){if(nodeP == NULL)return;clear(nodeP->left);clear(nodeP->right);delete nodeP;}BSTNode* get_root() const { return root; }bool isEmpty() const { return root == NULL; }BSTNode* search(const int& d) const{BSTNode* p = search(d, root);return p;}BSTNode* search(const int d, BSTNode* p) const{while(p != NULL){if(d == p->data)return p;else if(d < p->data)p = p->left;elsep = p->right;}return NULL;}void insert(const int d){BSTNode *p = root, *prev = NULL;while(p != NULL){prev = p;if(d < p->data)p = p->left;elsep = p->right;}if(root == NULL)root = new BSTNode(d);else if(d < prev->data)prev->left = new BSTNode(d);elseprev->right = new BSTNode(d);} };class linkedHash { private:BST* ht_bstree; //散列表二叉樹數(shù)組int bucket; //散列表桶個數(shù) public:linkedHash(int m):bucket(m){ht_bstree = new BST [bucket] ();}~linkedHash(){for(int i = 0; i < bucket; ++i){ht_bstree[i].clear(ht_bstree[i].get_root());}delete [] ht_bstree;}int hash(const int &key) const{return abs((((key+1000000000)%bucket)+1357)%bucket); //留余數(shù)法}int find(const int &x) const{int i = hash(x);BSTNode *p = ht_bstree[i].search(x);if(p)return p->count;return 0;}void insert(const int x){int i = hash(x);BSTNode *p = ht_bstree[i].search(x);if(p)p->count++;elseht_bstree[i].insert(x);} }; int a[4001], b[4001], c[4001], d[4001]; int ab[4000*4000+1], cd[4000*4000+1]; //存儲a+b，c+d int main() {linkedHash ht(5000); //多次調(diào)整括號內(nèi)數(shù)值，超時或者內(nèi)存超限int line, k=0;cin >> line;for(int i = 0; i < line; ++i){cin >> a[i] >> b[i] >> c[i] >> d[i];}for(int i = 0; i < line; ++i){for(int j = 0; j < line; ++j){ab[k] = a[i]+b[j];ht.insert(ab[k]);cd[k++] = -(c[i]+d[j]);}}int result = 0;for(int i = 0; i < k; ++i){result += ht.find(cd[i]);}cout << result << endl;return 0; }

2.6 AC代碼（哈希+數(shù)組法）

/*** @description: poj 2785 哈希法，數(shù)組實現(xiàn)* @author: michael ming* @date: 2019/5/20 18:17* @modified by: */ #include <iostream> using namespace std; int a[4001], b[4001], c[4001], d[4001]; const int hashtablesize = 20000001;//保證一定的富裕，裝載因子0.75左右 int hasht[hashtablesize]; int count[hashtablesize]; int offset = 1000000000; // 該數(shù) > 2^29（加上他后，就沒有負(fù)數(shù)了，求模后比較方便使用下標(biāo)值） int hashfunc(int value) {return value%hashtablesize; } int hashfunc_other(int value) {return (value+3)%hashtablesize; } void insert(int num) {int num_init = num;num = hashfunc(num+offset);while(hasht[num] != offset && hasht[num] != num_init)//解決沖突，不等于初始值（夠不著的大數(shù)）（值改了，位子被占了），且不等于映射的值（沖突了），第一次進(jìn)入循環(huán)，第一個條件肯定不滿足。{num = hashfunc_other(num);//沖突了，繼續(xù)尋找別的下標(biāo)（換一個函數(shù),不然相同的模在這可能無限循環(huán)）}hasht[num] = num_init;count[num]++; } int find(int num) {int num_init = num;num = hashfunc(num+offset);while(hasht[num] != offset && hasht[num] != num_init)num = hashfunc_other(num); //往下查找空位或者相等的值得位子if(hasht[num] == offset) //找到的是空位子，則沒有匹配的和等于0return 0;else //找到了值相等的位子，把其對應(yīng)的count位子里的個數(shù)返回return count[num]; } int main() {int line, k=0, value, i, j;cin >> line;for(i = 0; i < line; ++i){cin >> a[i] >> b[i] >> c[i] >> d[i];}for(i = 0; i < hashtablesize; ++i)hasht[i] = offset; //hash表每個元素初始化為offsetfor(i = 0; i < line; ++i){for(j = 0; j < line; ++j){value = a[i]+b[j];insert(value);}}int result = 0;for(i = 0; i < line; ++i){for(j = 0; j < line; ++j){value = (-c[i]-d[j]);result += find(value);}}cout << result << endl;return 0; }

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