skyline

#include #include char a[1000],b[2],c[1000];int judge(char *p){ int l=strlen(p); int i; for(i=0;i<l&&p[i]=='0';i++); p=&p[i]; l=strlen(p); if(l==10) return strcmp(

http://acm.hdu.edu.cn/showproblem.php?pid=3357按所给的顺序，给出一条有向边，如果加上这条边会出现环，那就舍弃这条边，问一共有几条边要舍弃如果加入的边a到b不会出现环，要做一次更新，有三类更新：i->a->b->j => i->j;i->a->b => i->b;a->b->j => a->j#include #include

http://acm.hdu.edu.cn/showproblem.php?pid=4287#include#include#include#include#includeusing namespace std;int alp[29]={2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9};int dic[12222222];

小明系列故事——师兄帮帮忙Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2603    Accepted Submission(s): 654Problem Description　　小明自从告别了ACM/

#include #include #include #define MAX 10using namespace std;struct matrix{ int num[MAX][MAX]; matrix() { memset(num,0,sizeof(num)); }};matrix res;matrix A;int m;mat

How many ways??Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 822    Accepted Submission(s): 254Problem Description春天到了, HDU校园里

#include #include #include #define MAX 15using namespace std;struct matrix{ int num[MAX][MAX]; matrix() { memset(num,0,sizeof(num)); }};matrix res;matrix A;int n,k;m

FibonacciTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6946 Accepted: 4908DescriptionIn the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn

#include #include #include #define MAX 35using namespace std;struct matrix{ int num[MAX][MAX]; matrix() { memset(num,0,sizeof(num)); }};int n;long long map[MAX][MAX];

Problem DescriptionOne day, ZHJ came across a warrior novel named Kunlun in library. Soon, ZHJ fall into it.He tells everybody how deeply he loves Kunlun no matter whom he meets.He found it at

这题花了很长时间，代码还是有点乱~~#include#include#include#includeusing namespace std;#define INF 10000000char map[105][105];int d[105][105];int visit[105][105];int xx[4]={1,-1,0,0};int yy[4]={0,0,1,-1};

其实就是八进制换10进制http://acm.hdu.edu.cn/showproblem.php?pid=4278#include#include#include#includeusing namespace std;int main(){ int n; char num[15]; while(scanf("%s",num)) { if(num[0]=='0'

http://acm.hdu.edu.cn/submit.php?pid=1058思路想到了，但是不会处理～～～～～～～～看了别人的佩服额～#include#include#include#includeusing namespace std;long long num[60000];int b[4]={2,3,5,7};long long min(long long a,

#include#include#define max 1000#define V 1000int c[max],w[max];long long dp[V][max]; //dp[V][N]表示剩余V空间，前N个物品的最大价值；int main(){ int n,v,i,j,o; scanf("%d",&o); while(o--) { scanf(

http://acm.hdu.edu.cn/showproblem.php?pid=1425#include#include#includeint a[1000001];int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF) { memset(a,0,sizeof(a));

#include#includeint a[10000];bool f[10000];int find(int t){ while(t!=a[t]) { t=a[t]; } return t;}int main(){ int t1,t2; int i; for(i=1;i<=10000;i++)

#include#includeusing namespace std;struct node { int i,j; int s;};bool cmp(node a,node b){ return a.s<b.s;}node a[1000000];int pa[1000000];int find(int t){

Doing Homework againTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3621    Accepted Submission(s): 2086Problem DescriptionIgnat

#include #include #include using namespace std;int tree1[10000];int tree2[10000];void Insert(char word,int *tree){ int now=1; int c=word-'0'; while(tree[now]!=-1) { if(t

思路是用递归~~ 比如1+3+4*7+3 先计算1+3   4+4*7+3 再  4+28+3 再 32+3 再35 ~~具体见代码#include#include#includevoid compute(double *dig,char *sign,int s1,int s2){ double temp1,temp2; if(sign[s2]=='*') {

http://acm.hdu.edu.cn/showproblem.php?pid=1053主要利用哈夫曼编码原理，再加广度有限搜索，思想想出来了～可BUG找了好长时间！！用队列的时候不能用node ,要用node *见下，还要注意字符串全部相同的情况;#include #include #include #include using namespace std;struct

http://acm.hdu.edu.cn/showproblem.php?pid=1027此题没什么算法，就是考组合数学吧！细节想了好久～/************************************************************************* > File Name: main.cpp > Author: huangshuai > Mai

分析后就是求树的直径两次bfs即可证明： 树的直径是指树的最长简单路。求法: 两遍BFS :先任选一个起点BFS找到最长路的终点，再从终点进行BFS，则第二次BFS找到的最长路即为树的直径；              原理: 设起点为u,第一次BFS找到的终点v一定是树的直径的一个端点              证明: 1) 如果u 是直径上的点，则v显然是直径的终点(因为如果v不