acm_算法模板_适合初学者使用.doc

1、1三角形面积计算 1字典树模板 2求线段所在直线 5求外接圆 5求内接圆 6判断点是否在直线上 8简单多边形面积计算公式 8stein 算法求最大共约数 .9最长递增子序列模板 o(nlogn 算法实现) .9判断图中同一直线的点的最大数量 10公因数和公倍数 12已知先序中序求后序 12深度优先搜索模板 13匈牙利算法 二部图匹配 BFS 实现 15带输出路径的 prime 算法 17prime 模板 .18kruskal 模板 .19dijsktra22并查集模板 23高精度模板 24三角形面积计算/已知三条边和外接圆半径，公式为 s = a*b*c/(4*R)double GetArea

2、(double a, double b, double c, double R)return a*b*c/4/R;/已知三条边和内接圆半径，公式为 s = prdouble GetArea(double a, double b, double c, double r)return r*(a+b+c)/2;/已知三角形三条边，求面积double GetArea(doule a, double b, double c)double p = (a+b+c)/2;return sqrt(p*(p-a)*(p-b)*(p-c);2/已知道三角形三个顶点的坐标struct Pointdouble x, y

3、;Point(double a = 0, double b = 0)x = a; y = b;double GetArea(Point p1, Point p2, Point p3)double t = -p2.x*p1.y+p3.x*p1.y+p1.x*p2.y-p3.x*p2.y-p1.x*p3.y+p2.x*p3.y;if(t #include #include #define BASE_LETTER a#define MAX_TREE 35000#define MAX_BRANCH 26structint nextMAX_BRANCH; /记录分支的位置int cMAX_BRANCH;

4、 /查看分支的个数int flag; /是否存在以该结点为终止结点的东东，可以更改为任意的属性 trieMAX_TREE;int now;void init()now = 0;memset(3now +;int add ()memset(return now+;int insert( char *str)int pre = 0, addr;while( *str != 0 )addr = *str - BASE_LETTER;if( !triepre.nextaddr )triepre.nextaddr = add();triepre.caddr+;pre = triepre.nextaddr

5、;str +;triepre.flag = 1;return pre;int search( char *str )int pre = 0, addr;while( *str != 0 )addr = *str - BASE_LETTER;if ( !triepre.nextaddr )return 0;pre = triepre.nextaddr;str +;if( !triepre.flag )return 0;4return pre;pku2001 题，源代码：void check( char *str )int pre = 0, addr;while(*str != 0)addr =

6、*str - BASE_LETTER;if( triepre.caddr = 1)printf(“%cn“, *str);return;printf(“%c“, *str);pre = triepre.nextaddr;str +;printf(“n“);char input100125;int main()int i = 0,j;init();while(scanf(“%s“, inputi) != EOF)getchar();insert(inputi);i+;for(j = 0; j p1.x int main()int count,i,n,left,mid,right,Blen=0,n

7、um;scanf(“%d“, /case 的个数while(count-)scanf(“%d“, /每组成员的数量Blen = 0;for(i=1;i#include #include using namespace std;#define MAX 1010 /最大点的个数struct pointint x,y;numMAX;int usedMAXMAX*2; /条件中点的左边不会大于 1000,just equal MAXint countNMAXMAX*2;#define abs(a) (a0?a:(-a) int GCD(int x, int y)int temp;if(x #inclu

8、de using namespace std;string post;void fun(string pre, string mid)13if(pre = “ | mid = “) return;int i = mid.find(pre0);fun(pre.substr(1,i), mid.substr(0,i);fun(pre.substr(i+1, (int)pre.length()-i-1), mid.substr(i+1, (int)mid.length()-i-1);post += pre0;int main()string pre, mid;while(cin pre)cin mi

9、d;post.erase();fun(pre, mid);cout = 1 search(x-1,y); if(x+1 = 1 search(x,y-1);if(y+1 = 1)if(datax-1y = t)count+;search(x-1,y);if(y-1 = 1 search(x-1,y-1);if(y+1 = 1 15search(x+1,y-1);if(y+1 = 1 search(x,y-1);if(y+1 -1 )17 vm1vm2prevx = x; vm2x = vm2prevx; x = prevx; vm2x = i; vm1i = x; /匹配的边数加一 ans+;

10、 return ans; 带输出路径的 prime 算法#include #include using namespace std;const int MAX = 110;int dataMAXMAX;int lowcostMAX;int adjvexMAX;int main()int n;cin n;int i,j;for(i = 0 ;i dataij;/primfor(i = 1; i %dn“, adjvexchoose+1, choose+1, lowcostchoose);lowcostj = 0;for(j = 1; j datachoosej)lowcostj = datach

11、oosej;adjvexj = choose;return 0;prime 模板#include #include #include using namespace std;int const MAX = 110;int disMAXMAX;int lowcostMAX;int main()int n;int i,j;while(cin n)for(i = 0; i disij;/下面是 prim 算法部分，ans 是计算所有路径的和lowcost0 = 0;for(i = 1; i dischoosej)lowcostj = dischoosej;cout #include #include

12、 using namespace std;20const int MAX = 1010; /节点个数const int MAXEDGE = 15010; /边个数bool usedMAXEDGE; /标记边是否用过struct nodeint begin, end, dis;dataMAXEDGE;class UFSetprivate:int parentMAX+1;int size;public:UFSet(int s = MAX);int Find(int x);void Union(int root1, int root2);UFSet:UFSet(int s)size = s+1;me

13、mset(parent, -1, sizeof(int)*size);void UFSet:Union(int root1, int root2)int temp = parentroot1 + parentroot2;if(parentroot1 0)p = parentp;21int t = x;while(t != p)t = parentx;parentx = p;x = t;return p;bool cmp(node a, node b)return (a.dis #include using namespace std;const int maxint = 9999999;con

14、st int maxn = 1010;int datamaxnmaxn, lowcostmaxn; /data 存放点点之间的距离，lowcost 存放点到 start 的距离,从 0 开始存放bool usedmaxn;/标记点是否被选中int n; /顶点的个数void disktra(int start)/初始点是 start 的 dij 算法int i,j;memset(used, 0, sizeof(used);/initefor(i = 0; i lowcostj)choose = j;tempmin = lowcostj;usedchoose = true;23/updata o

15、thersfor(j = 0; j #include using namespace std;const int MAX = 5005;class UFSetprivate:int parentMAX+1;int size;public:UFSet(int s = MAX);/初始化void Union(int root1, int root2);/合并，注意参数为根节点int Find(int i );/返回根节点int SetNum();/返回集合的个数;UFSet:UFSet(int s)size = s+1;memset(parent, -1, sizeof(int)*size );v

16、oid UFSet:Union( int root1, int root2)int temp = parentroot1+parentroot2;if(parentroot1=0; j = parentj);while(i!=j)int temp = parenti;parenti = j; i = temp;return j;int UFSet:SetNum()int totalNum = 0,i;for(i = 0; i #include using namespace std;/ - 非负数计算部分：f1f14 -string operator+(string x, string y);

17、 / x、y 都必须非负string operator-(string x, string y); / x、y 都必须非负（结果可能为负）string operator*(string x, string y); / x、y 非负string operator*(string s, int a); / s,a 非负，且 a 必须小于 2*108.string MSDiv(string x, int y, int / 多精度除以 int，x 非负，y为正string operator/(string s, int a); / 调用 MSDiv25int operator%(string s, i

18、nt a); / 调用 MSDivstring MMDiv(string x, string y, string / 多精度除以多精度，x 非负，y 为正string operator/(string x, string y); / 调用 MMDivstring operator%(string x, string y); / 调用 MMDivstring HPower(string s, int a); / s,a 必须非负！string HSqrt(string s); / 开平方取整，s 非负！string Head_zero_remover(string num); / 除了开头可能有

19、0外，num 必须是非负数。bool Less(string x, string y); / 非负数之间的“小于”/ - 以下是负数支持：f15f19 -string operator-(string s); / 取负string SAdd(string x, string y);string SMinus(string x, string y);string SMul(string x, string y);string SMul(string s, int a); / 同样，a 的绝对值不能超过 2*108/ - f1 () -string operator+(string x, stri

20、ng y)if(x.size() =yx.swap(y);y.insert(y.begin(), x.size()-y.size(), 0); / y 开头补 0 到和 x 一样长string sum(x.size(), -1); / 初始大小：x.size()int carry=0;for(int i=x.size()-1; i = 0; -i)carry += xi+yi-2*0;sumi = carry%10+0;carry /= 10;if(carry 0) / 还有进位 1return string(“1“) += sum; / 给开头添加一个“1”return sum;26/ -

21、f2 (need: f13, f14) -string operator-(string x, string y)bool neg = false; / 结果为负标志if(Less(x, y)x.swap(y); / 如果 x= 0; -i)if(xi = yi+carry) / 本位够减diffi = xi-yi-carry+0;carry = 0;else / 需要借位diffi = 10+xi-yi-carry+0;carry=1;if(neg)return string(“-“) += Head_zero_remover(diff);return Head_zero_remover(d

22、iff); / - f3 (need f1, f4) -string operator*(string x, string y)string prod=“0“; /初值：0for(int i=y.size()-1; i = 0; -i)string p_sum = x * (yi-0); / p_sum：部分积if(p_sum != “0“) / 保证后面加 0 后也符合 UAdd 的要求！p_sum.insert(p_sum.end(), y.size()-1-i, 0);prod = prod + p_sum; 27return prod;/ - f4 () -string operato

23、r*(string s, int a)if(s = “0“ | a = 0) / 以免后面特殊处理！return “0“; string prod(s.size(), -1); / 先申请 s.size()位int carry=0;for(int i=s.size()-1; i = 0; -i)carry += (si-0)*a;prodi = carry%10+0;carry /= 10;while(carry0)prod.insert(prod.begin(), carry%10+0);carry /= 10;return prod;/ - f5 (need f13) -string MS

24、Div(string x, int y, int res=0;for(int i=0; i0)if(a%2 = 1)power = power * s;a /= 2;s = s * s;return power;/ - f12 (need f2, f4, f13, f14) -string HSqrt(string s) / 手工开平方。若要返回余数，return 前的 res 就是！string sqroot(s.size()+1)/2, -1);string res = s.substr(0, 2-s.size()%2); / 奇位取前 1，偶位取前 2string div=“0“; /

25、占一位置for(int i=0; isqroot.size(); +i)for(int quot=9; ; -quot)divdiv.size()-1 = quot+0; / 末位试商，从9到0string p_prod = div*quot;if( ! Less(res, p_prod) ) / p_prod = ressqrooti = quot+0; / 将结果追加！div = sqroot.substr(0, i+1)*20;res = res - p_prod;30string next2 = s.substr(i+1)*2-s.size()%2, 2);if(res = “0“)r

26、es = next2; / 取后 2 位elseres += next2; / 下移 2 位，追加；即 res = res*100+next2break;return sqroot;/ - f13 () -bool Less(string x, string y)return x.size()y.size() | x.size() = y.size() / - f14 () -string Head_zero_remover(string num) / 化简“003”等数if(num0 != 0)return num;int pos=num.find_first_not_of(0);if(pos = string:npos) / 全 0return “0“;return num.substr(pos, num.size()-pos);/ 以下是负数支持！/ - f15 () -string operator-(string s)if(s0 = -)return s.substr(1, s.size()-1);if(s = “0“)