分治思想《算法竞赛入门经典》

三步法
1，划分问题
2，递归求解
3，合并问题

首相分治法可以用了优化最大连续和
优化后时间复杂度为o（nlogn)

第一步：利用m=x+(y-x)/2，设置划分点。注意划分格式为[x,m),[m,y),这样不仅符合左边去到右边取不到的规范，而且不用m-1，m+1这些容易出错的步骤。
第二步:划分后近似相当于T(n)=2(n/2),然后对两边递归求解，最小不可分元素就是当一段连续和中只有一个元素，所以当right-left=1时返回A[left],因为left能取到，right取不到。当不是最小元素时，就左右两边延长到底，和该段每个元素单个值比较，留下最大的，返回给上一层。

棋盘覆盖问题
递归至边长只有2的正方形，此时必有一块已经被填，只需要将剩下三块填完即可，大于两块将棋盘一分为四，有个象限含有初始的黑块，将中心四块中其它的三块涂黑，然后继续递归。

#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<algorithm>#include<iostream>#include<time.h>#include<map>using namespace std;const int INF = 0x3f3f3f3f;int board[105][105];int team = 1;int pos(int size, int a, int b, int x, int y){    if (x < a + size / 2 && y < b + size / 2)        return 1;    else if (x >= a + size / 2 && y < b + size / 2)        return 2;    else if (x >= a + size / 2 && y >= b + size / 2)        return 3;    else        return 4;}void Divide(int size, int a, int b){    int x, y;    if (size == 2)    {        for (int i = a; i <a + size; i++)        {            for (int j = b; j <b + size; j++)            {                if (board[i][j] == -1)                {                    board[i][j] = team;                }            }        }        team++;    }    else    {        for (int i = a; i <a + size; i++)        {            for (int j = b; j <b + size; j++)            {                if (board[i][j] != -1)                {                    x = i;                    y = j;                }            }        }        int loc = pos(size, a, b, x, y);        for (int i = a + size / 2 - 1; i <= a + size / 2; i++)        {            for (int j = b + size / 2-1; j <= b + size / 2; j++)            {                if (loc != pos(size, a, b, i, j))                    board[i][j] = team;            }        }        team++;        Divide(size / 2, a, b);        Divide(size / 2, a + size / 2, b + size / 2);        Divide(size / 2, a + size / 2, b);        Divide(size / 2, a, b + size / 2);    }}int main(){    int k;    while (cin >> k, k)    {        int x, y;        cin >> x >> y;        memset(board, -1, sizeof(board));        board[x][y] = 0;        int size = 1;        for (int i = 1; i <= k; i++)        {            size *= 2;        }        Divide(size, 1, 1);        for (int i = 1; i <= size; i++)        {            for (int j = 1; j <= size; j++)            {                printf("%-4d", board[i][j]);            }            printf("\n");        }    }    return 0;}

循环日程表问题
思路通过temp控制每次方格的大小，来实现分治，
有规律可循，通过规律先对左下赋值
然后将左下复制到右上，
再将左上复制到右下

#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<algorithm>#include<iostream>#include<time.h>#include<map>using namespace std;const int INF = 0x3f3f3f3f;int board[105][105];void solve(int k){    board[1][1] = 1;    board[2][1] = 2;    board[1][2] = 2;    board[2][2] = 1;    int  temp = 1;    for (int s = 1; s <= k; s++)    {        temp *= 2;        for (int i = 1; i <= temp;i++)        {            for (int j = 1; j <=temp;j++)            {                board[i + temp][j] = board[i][j] + temp;            }        }        for (int i = 1; i <= temp; i++)        {            for (int j = 1; j <=temp; j++)            {                board[i][j + temp] =board[i + temp][j] ;            }        }        for (int i = 1; i <= temp; i++)        {            for (int j = 1; j <=temp; j++)            {                board[i+temp][j + temp] =board[i][j] ;            }        }       }}int main(){    int k;    while (cin >> k, k)    {        solve(k);        int size;        size = 1;        for (int i = 1; i <= k; i++)        {            size *= 2;        }        for (int i = 1; i <= size; i++)        {            for (int j = 1; j <= size; j++)            {                printf("%-4d", board[i][j]);            }            printf("\n");        }    }    return 0;}

巨人与鬼
题目大意是n个黑点，n个白点，一个黑点连一个白点（直线），如何使任意两条直线都没有相交。

#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<algorithm>#include<iostream>#include<time.h>#include<map>using namespace std;const int INF = 0x3f3f3f3f;int n;struct node{    int x, y;    int type;    int ID;    double angle;}points[1005];bool cmp1(node a, node b){    if (a.y == b.y)        return a.x < b.y;    return a.y < b.y;}bool cmp2(node a, node b){    return a.angle < b.angle;}void swap(int temp1, int temp2){    int temp;    temp = temp1;    temp1 = temp2;    temp2 = temp;}int ans[10005];void Swap(node a, node b){    swap(a.x, b.x);    swap(a.y, b.y);    swap(a.ID, b.ID);    swap(a.type, b.type);    double temp;    temp = a.angle;    a.angle = b.angle;    b.angle = temp;}void solve(int left, int right){    if (right - left == 1)    {        ans[points[left].ID] = points[right].ID;        return;    }    if (left >= right)    {        return;    }    sort(points + left, points + right + 1, cmp1);    for (int i = left+1; i <= right; i++)    {        points[i].angle = atan2(points[i].y - points[left].y, points[i].x - points[left].x);    }    sort(points + left+1, points + right + 1, cmp2);    int sum = 0;    for (int i = left; i <= right; i++)    {        sum += points[i].type;        if (sum == 0)        {            ans[points[left].ID] = points[right].ID;            solve(left + 1, i - 1);            solve(i + 1, right);        }           }}int main(){    while (cin >> n, n)    {        for (int i = 1; i <= n*2; i++)        {            cin >> points[i].type;            cin >> points[i].x;            cin >> points[i].y;            points[i].ID = i;        }        solve(1, 2*n);        for (int i = 1; i <=n; i++)            printf("%d\n", ans[i]);    }    return 0;}