月度归档:2017年10月

[POJ 3617]Best Cow Line

Description

FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual”Farmer of the Year” competition. In this contest every farmer arranges his cows in a line and herds them past the judges.

The contest organizers adopted a new registration scheme this year: simply register the initial letter of every cow in the order they will appear (i.e., If FJ takes Bessie, Sylvia, and Dora in that order he just registers BSD). After the registration phase ends, every group is judged in increasing lexicographic order according to the string of the initials of the cows’ names.

FJ is very busy this year and has to hurry back to his farm, so he wants to be judged as early as possible. He decides to rearrange his cows, who have already lined up, before registering them.

FJ marks a location for a new line of the competing cows. He then proceeds to marshal the cows from the old line to the new one by repeatedly sending either the first or last cow in the (remainder of the) original line to the end of the new line. When he’s finished, FJ takes his cows for registration in this new order.

Given the initial order of his cows, determine the least lexicographic string of initials he can make this way.

Input

  • Line 1: A single integer: N
  • Lines 2~N+1: Line i+1 contains a single initial (‘A’..’Z’) of the cow in the ith position in the original line

Output

The least lexicographic string he can make. Every line (except perhaps the last one) contains the initials of 80 cows (‘A’..’Z’) in the new line.

Sample Input

6
A
C
D
B
C
B

Sample Output

ABCBCD

Hint

使用贪心算法,不断将较小的字符放入新字符串的末尾,这样我们就能构造一个字典序最小的串。对于首尾字符相同的情景,取哪个都可以。
注意题目要求,每 80 字符换行。

题解

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <cstring>
//#include <regex>
#define online
#define INF 2010
using namespace std;

int n, head, tail;
char str[INF];
string str2;
int chk_head,chk_tail;
inline bool check()
{
    #ifdef offline
    printf("new check\n");
    #endif
    chk_head = head;
    chk_tail = tail;
    while (chk_head <= chk_tail)
    {
        if (str[chk_head] == str[chk_tail])
        {
            #ifdef offline
            printf("str[chk_head] %d %c== str[chk_tail] %d %c\n", chk_head, str[chk_head], chk_tail, str[chk_tail]);
            #endif
            chk_head++;
            chk_tail--;
        }
        else
        {
            if (str[chk_head] > str[chk_tail])
            {
                str2 = str2 + str[tail];
                tail--;
                #ifdef offline
                printf("check end at tail\n");
                #endif
                return 1;//tail
            }
            else
            {
                str2 = str2 + str[head];
                head++;
                #ifdef offline
                printf("check end at head\n");
                #endif
                return 0;//head
            }
        }
    }
    str2 = str2 + str[tail];
    tail--;
    #ifdef offline
    printf("check end at outside\n");
    #endif
}
inline bool solve()
{
    #ifdef offline
    printf("new solve\n");
    #endif
    if (str[head] > str[tail]){
        str2 = str2 + str[tail];
        tail--;
        #ifdef offline
        printf("solve 1\n");
        #endif
    }
    else
    {
        #ifdef offline
        printf("solve 0\n");
        #endif

        if (str[head] == str[tail] && head != tail){
            #ifdef offline
            printf("head == tail ==  %d\n", head);
            #endif
            check();
        }
        else
        {
            str2 = str2 + str[head];
            head++;
        }
    }
    if (head > tail)
        return 0;
    else
        solve();
}

int main()
{
    //srand((unsigned int)(time(NULL)));
    //freopen("","r",stdin);
    //freopen("","W",stdout);
    scanf("%d", &n);
    head = 0;
    tail = n-1;
    for (int i = 0; i < n; i++)
        scanf("%c%c", &str[i], &str[i]);
    solve();
    //cout<<str2;//poisonous output format
    for (int i = 0; i < str2.length(); i++)
    {
        if ((i + 1) % 80 == 0){
            if (i + 1 == n)
                printf("%c", str2[i]);
            else
                printf("%c\n", str2[i]);
        }
        else
            printf("%c", str2[i]);
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

Dijkstra 最短路径算法

概要

Dijkstra 算法是最常用的求最短路径的算法之一,它同时适用于有向图或无向图。
我们用 w[u][v] 来存储节点 u 到节点 v 的距离(或理解为这条边的权重)。将除源 S 以外的节点分成已标记和未标记的两部分,先更新一次源 S 到相邻节点的距离,然后通过 BFS(广度优先搜索)来遍历所有与未标记的与源 S 间接相连的节点并不断维护从源 S 到 i点的距离 dis[i],最终得到最短路径。这个版本的 Dijkstra 最短路径算法的复杂度约为 O(N^2)。

输入

  • 第 1 行 节点个数 n,边个数 m
  • 第 2~m+1行 边起点 ui,边终点 vi,边权重 w[u][v]

实现

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <cstring>
#define offline
//#include <regex>
#define INF 2147483647
using namespace std;
int main()
{
    //srand((unsigned int)(time(NULL)));
    freopen("Dijkstra.in","r",stdin);
    int n, m, u;//n points,m threads, u as temp source
    scanf("%d%d", &n, &m);
    int w[n][n], d[n];

    bool sign[n];
    memset(sign,0,sizeof(sign));
    sign[0] = 1;
    
    d[0] = 0;
    for (int i = 0; i < n; i++)
        d[i] = INF;

    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            i == j ? w[i][j] = 0: w[i][j] = INF;

    int t1, t2, t3;
    for (int i = 0; i < m; i++)
    {
        scanf("%d%d%d", &t1, &t2, &t3);
        w[t1][t2] = t3;
        w[t2][t1] = t3;//only for non-direction map
    }
    //initial for d
    for (int i = 0; i < n; i++)
        d[i] = w[0][i];
    
    
    int min;
    for (int i = 0; i < n; i++)
    {
        min = INF;
        for (int j = 0; j < n; j++)
        {
            if (sign[j] == 0 && d[j] < min)
            {
                min = d[j];
                u = j;
            }
            
        }
        sign[u] = 1;
        for (int v = 0; v < n; v++)
        {
            if ((d[v] > d[u] + w[u][v]) && w[u][v] < INF){
                #ifdef offline
                printf("updated d[%d] from %d to %d\n", v,d[v],d[u]+w[u][v]);
                #endif
                d[v] = d[u] + w[u][v];
            }
        }
    }   
    for (int i = 0; i < n; i++)
        printf("dis[%d] %d\n", i, d[i]);
    return 0;
}

[POJ 2386]Lake Counting

Description

有一个大小为 N × M 的园子,雨后积起了水。八连通的积水被认为是连接在一起的。请求出
园子里总共有多少水洼?(八连通指的是下图中相对 W 的*的部分)

* * *
* W *
* * *

Input

第 1 行 两个数 N, M
第2~n行 各 M 个数

Sample Input

10 12
W........WW. 
.WWW.....WWW 
....WW...WW. 
.........WW. 
.........W.. 
..W......W.. 
.W.W.....WW. 
W.W.W.....W. 
.W.W......W. 
..W.......W.

Sample Output

3

题解

通过二维布尔表来记录积水,搜索邻接坐标的状态并重置以避免重复计数。

#include "cstdio"
#include "cmath"
#include "cstdlib"
#include "iostream"
#include "cstring"
#define INF 110
#define online
using namespace std;
int count = 0;
int n,m;
bool map[INF][INF];
inline bool handle(int n1,int m1,bool sign)
{
    //endowed W as 1, NULL and * as 0
    //use sign to distinguish the original requestion and derived requestion
    //True->Origin
    if (n1 < 0 || m1 < 0 || n1 >= n || m1 >= m)
    {
        #ifdef debug
        printf("n1 %d m1 %d n %d m %d out\n", n1, m1, n, m);
        #endif
        return 0;
    }
    if (map[n1][m1])
    {
        #ifdef debug
        printf("matched\n");
        #endif
        map[n1][m1] = 0;
        if (sign)
            count++;
        handle(n1,m1-1,0);
        handle(n1,m1+1,0);
        handle(n1+1,m1,0);
        handle(n1-1,m1,0);
        handle(n1-1,m1-1,0);
        handle(n1+1,m1+1,0);
        handle(n1+1,m1-1,0);
        handle(n1-1,m1+1,0);
    }

}

int main()
{
    //initialize
    scanf("%d%d", &n, &m);
    char c, str[m+1];
    memset(map,0,sizeof(map));
    //endowed * as 0, W as 1
    for (int i = 0; i < n; i++)
    {
        scanf("%s", &str);
        for (int j = 0; j < m; j++)
            map[i][j] = str[j] == '.' ? 0 : 1;

    }

    #ifdef debug
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++)
            printf("%d", map[i][j]);
        printf("\n");
    }
    #endif

    //handle
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            handle(i,j,1);
    printf("%d\n", count);

    return 0;
}

C 编程求出变量类型的内存空间长度

题目

编程求出 char、short、int、long、long long、float、double、long double 变量类型的内存空间长度。

题解

sizeof 运算符

sizeof(类型)

返回 类型 的对象表示的字节数

sizeof 表达式

返回当 表达式 求值时所返回的类型的对象的表示的字节数

引用自 sizeof 运算符

示例代码如下,其余的可以以此类推。
更正:由于 sizeof 是数据类型空间长度计算符号,所以可以直接求出变量类型的数据类型空间长度无需单独定义变量。修改后的题解如下。(感谢上理计科 夏耘 老师指正!)

printf("%u\n", sizeof(int));

原题解如下:

int a;
printf("%u\n", sizeof(a));

注:sizeof() 返回值为 size_t 类型,而 size_t 的类型和编译器有关,是不确定的。唯一可以确定的是这是个 unsigned 类型。在此处我们使用 %u 来输出。(C99 和 C11 提供 %zd 转换说明匹配 size_t,一些不支持 C99 和 C11 的编译器可用 %u%lu 代替。)

二进制思想

我们已知:1 byte = 8 bits。又知,对于题目中所述的变量类型,一般通过一个单独的 bit 来表示数值的正负,为 0 时表示正,为 1 时表示负。这就意为着,当所有的 bit 全为 1 时,值为负数且绝对值最大。那么我们可以构造一个循环来计算 2 的 n 次幂并存入变量,当完成最后一次循环时,最高位比特为 1,数值溢出变量范围,变为负值。此时 n+1 为该变量类型内存空间所占比特数。通过关系 1 byte = 8 bits 即可求得对应的字节数。 示例代码如下,其余的可以以此类推。注:因为完成了最后一次循环后 i 进行了自增,因此实际输出的 n 即为刚刚我们所提到的 n+1。

char t = 1;
int i = 1;
for (; t > 0; i++)
    t *= 2;
printf("%d\n", i);