哇，果然是一曝十寒啊，这么多天没总结了。
再次整理了一下同余模问题：
附上简单入门题：
http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1001&cid=781
A Math Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1867 Accepted Submission(s): 519
Problem Description
You are given a positive integer n, please count how many positive integers k satisfy kk≤n.
Input
There are no more than 50 test cases.
Each case only contains a positivse integer n in a line.
1≤n≤1018
Output
For each test case, output an integer indicates the number of positive integers k satisfy kk≤n in a line.
Sample Input
1
4
Sample Output
1
2
#include <iostream>
#include <cstring>
#define LL long long;
LL kpow(int x, int n){
LL res = 1;
while(n >= 0){
if(n & 1) res = res * x;
x = x * x;
n >>= 1;
}
return res;
}
int main(){
LL n;
while(~scanf("%lld",&n)){
for(int k = 15; k >= 1; k--){
if(kpow(k,k) <= n){
printf("%d\n",k);
break;
}
}
}
return 0;
}
今天人工智能考到A算法，错的惨兮兮
回来补补课
POJ 2243 一道A入门题
Knight Moves
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 14679 Accepted: 8226
Description
A friend of you is doing research on the Traveling Knight Problem (TKP) where you are to find the shortest closed tour of knight moves that visits each square of a given set of n squares on a chessboard exactly once. He thinks that the most difficult part of the problem is determining the smallest number of knight moves between two given squares and that, once you have accomplished this, finding the tour would be easy.
Of course you know that it is vice versa. So you offer him to write a program that solves the "difficult" part.
Your job is to write a program that takes two squares a and b as input and then determines the number of knight moves on a shortest route from a to b.
Input
The input will contain one or more test cases. Each test case consists of one line containing two squares separated by one space. A square is a string consisting of a letter (a-h) representing the column and a digit (1-8) representing the row on the chessboard.
Output
For each test case, print one line saying "To get from xx to yy takes n knight moves.".
Sample Input
e2 e4
a1 b2
b2 c3
a1 h8
a1 h7
h8 a1
b1 c3
f6 f6
Sample Output
To get from e2 to e4 takes 2 knight moves.
To get from a1 to b2 takes 4 knight moves.
To get from b2 to c3 takes 2 knight moves.
To get from a1 to h8 takes 6 knight moves.
To get from a1 to h7 takes 5 knight moves.
To get from h8 to a1 takes 6 knight moves.
To get from b1 to c3 takes 1 knight moves.
To get from f6 to f6 takes 0 knight moves.
#include <iostream>
#include <queue>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;
struct knight{
int x,y,step;
int g,h,f;
bool operator < (const knight &k) const{
return f > k.f;
}
}k;
bool visited[8][8];
int x2,y2,ans;
int dir[8][2] = {{-2,-1},{-2,1},{2,-1},{2,1},{-1,-2},{-1,2},{1,-2},{1,2}};
priority_queue<knight> que;
bool in (const knight &a){
if(a.x < 0 || a.y < 0 || a.x >= 8 || a.y >= 8){
return false;
}
return true;
}
//曼哈顿估价函数
int Heuristic(const knight & a){
return (abs(a.x - x2) + abs(a.y - y2)) * 10;
}
void Astar(){
knight t,s;
while(!que.empty()){
t = que.top();
que.pop();
visited[t.x][t.y] = true;
if(t.x == x2 && t.y == y2){
ans = t.step;
break;
}
for(int i = 0; i < 8; i++){
s.x = t.x + dir[i][0];
s.y = t.y + dir[i][1];
if(in(s) && !visited[s.x][s.y]){
s.g = t.g + 23;
s.h = Heuristic(s);
s.f = s.g + s.h;
s.step = t.step + 1;
que.push(s);
}
}
}
}
int main(){
char line[5];
int x1,y1;
while(gets(line)){
x1 = line[0] - 'a';
y1 = line[1] - '1';
x2 = line[3] - 'a';
y2 = line[4] - '1';
memset(visited,false,sizeof(visited));
k.x = x1;
k.y = y1;
k.g = k.step = 0;
k.h = Heuristic(k);
k.f = k.g + k.h;
while(!que.empty()) que.pop();
que.push(k);
Astar();
printf("To get from %c%c to %c%c takes %d knight moves.\n",line[0],line[1],line[2],line[3],ans);
}
return 0;
}
另外这题还可以用双向BFS解决，待补坑。。。
再挖一坑，最近碰到线段树这一新的数据结构，一道入门题练手 HDU 1166，挖坑待补。。。
四子连棋问题
BFS + 数码状态表示 + hash判重
#include <stdio.h>
#include <map>
#include <queue>
#include <stdlib.h>
using namespace std;
struct status
{
int step; //step=走的棋总步数
int hash; //哈希状态
int map[6][6]; //记录棋盘状态 map[i][j]=第i行第j列
int last; //last=上一回合下的棋的颜色,last=1表示黑棋，last=2表示白棋
}first; //first=没下棋之前的初始状态
queue <struct status> Q; //保存BFS每一层搜索结果的队列Q
map<int,bool>h[3]; //BFS判重map数组
int xx[]={1,-1,0,0},yy[]={0,0,1,-1};
int gethash(status tmp) //获取tmp棋盘的哈希值
{
int sum=0,i,j,k=1;
for(i=1;i<=4;i++)
{
for(j=1;j<=4;j++)
{
sum+=tmp.map[i][j]*k;
k*=3;
}
}
return sum;
}
int check(status tmp) //判断棋盘tmp是否达到目标要求,是则返回1，不是则返回0
{
int i,j,k,flag;
for(i=1;i<=4;i++) //检查每一横排是否有四连棋
{
flag=1;
for(j=1;j<4;j++)
if(tmp.map[i][j]!=tmp.map[i][j+1])
flag=0;
if(flag==1) return 1;
}
for(j=1;j<=4;j++) //检查每一竖列是否有四连棋
{
flag=1;
for(i=1;i<4;i++)
if(tmp.map[i][j]!=tmp.map[i+1][j])
flag=0;
if(flag==1) return 1;
}
if(tmp.map[1][1]==tmp.map[2][2]&&tmp.map[2][2]==tmp.map[3][3]&&tmp.map[3][3]==tmp.map[4][4]) //检查对角线上是否有四连棋
return 1;
if(tmp.map[1][4]==tmp.map[2][3]&&tmp.map[2][3]==tmp.map[3][2]&&tmp.map[3][2]==tmp.map[4][1]) //检查对角线上是否有四连棋
return 1;
return 0;
}
void move(status now,int x,int y,int k) //move(上一回合棋盘状态，棋盘空格x坐标，棋盘空格y坐标，移动方向k)
{
status tmp=now;
int tmpx=x+xx[k]; //tmpx=空格到达的目标棋格x坐标
int tmpy=y+yy[k]; //tmpy=空格到达的目标棋格y坐标
if(tmpx<1||tmpx>4) return; //目标棋格坐标越界，退出
if(tmpy<1||tmpy>4) return;
if(tmp.map[tmpx][tmpy]==tmp.last) return; //被移动的棋子不是我方颜色的，退出
tmp.last=3-tmp.last; //交换棋子颜色，这次下和上次不同的棋
swap(tmp.map[x][y],tmp.map[tmpx][tmpy]); //移动棋子
tmp.step++; //移动步数+1
tmp.hash=gethash(tmp);
if(check(tmp)) //移动后棋子到达目标状态,直接输出结果，退出程序
{
printf("%d\n",tmp.step);
exit(0);
}
if(!h[tmp.last][tmp.hash]) //该状态没有被访问过
{
h[tmp.last][tmp.hash]=1; //标记该状态访问过
Q.push(tmp); //将该状态的tmp入队
}
}
void bfs() //bfs广搜过程
{
first.hash=gethash(first);
first.last=1; //因为第一次下棋的首方不清楚，因此先将第一个是下黑棋的入队
Q.push(first);
first.last=2; //再将第一个是下白棋的入队
Q.push(first);
while(!Q.empty()) //当队列中还有棋盘状态没有搜索完(队列为空)，继续搜索
{
status now;
int x1=-1,x2=-1,y1=-1,y2=-1,i,j,k;
now=Q.front(); //取出队首的棋盘状态now
Q.pop();
for(i=1;i<=4;i++)
{
for(j=1;j<=4;j++)
{
if(now.map[i][j]==0)
{
if(x1==-1&&x2==-1) //找到第一个空格
{
x1=i;
y1=j;
}
else //否则，第一个空格已经找到，现在找到了第二个空格
{
x2=i;
y2=j;
}
}
}
}
//这步棋有两种选择：
//1、移动第1个空格
//2、移动第2个空格
//分别进行四个方向的移动、状态入队操作
for(k=0;k<4;k++)
{
move(now,x1,y1,k); //移动第1个空格
move(now,x2,y2,k); //移动第2个空格
}
}
}
int main()
{
int i,j;
for(i=1;i<=4;i++)
{
char in[10];
scanf("%s",in);
for(j=1;j<=4;j++)
{
if(in[j-1]=='B')
first.map[i][j]=1;
else if(in[j-1]=='W')
first.map[i][j]=2;
else if(in[j-1]=='O')
first.map[i][j]=0;
}
}
bfs();
return 0;
}
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