Codeforces Beta Round #3

A. Shortest path of the king

题意：在给定国际象棋上王的起始位置，求王到另一位置走的最少步数。并一种移动方案。王走的八个方向分别为用（L，R，U，D，LR，RD，LU，LD）表示

输入：给定起点和终点

输出：要走的最少路和一种走的方式。

题解：假设王的开始的坐标为（x0,y0）,终点为（x1,y1）,则容易得到下一步要尽量靠尽，所以要让(x,y)都有变化，即尽量走斜线。

代码：

//written by nothi//话说写麻烦了#include <math.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#define MAX(a,b) ((a)>(b)?(a):(b))struct Cood{    int x;    int y;};Cood begin, end;//L R U D LU LD RU RDint maxp = 8;const int dx[] = {-1,+1,0,0,-1,-1,+1,+1};const int dy[] = {0,0,+1,-1,+1,-1,+1,-1};void print(int i){    if(i == 0) printf("L");    if(i == 1) printf("R");    if(i == 2) printf("U");    if(i == 3) printf("D");    if(i == 4) printf("LU");    if(i == 5) printf("LD");    if(i == 6) printf("RU");    if(i == 7) printf("RD");    printf("\n");}int main(){    char a, b;    int i, j, x, y, ans;    scanf(" %c%d %c%d",&a, &x, &b, &y);    begin.x = a - 'a' + 1;    begin.y = x;    end.x = b - 'a' + 1;    end.y = y;    ans = MAX(abs(begin.x - end.x),abs(begin.y - end.y));    printf("%d\n",ans);    int cur_x = begin.x;    int cur_y = begin.y;    for(i = 0; i < ans; i++){int min = ans+1, now;for(j = maxp-1; j >= 0; j--){    int u = cur_x + dx[j];    int v = cur_y + dy[j];    int cur = MAX(abs(u - end.x),abs(v - end.y));    if(cur == ans - i - 1){min = cur;now = j;    }}print(now);cur_x += dx[now];cur_y += dy[now];    }    return 0;}

B. Lorry

暂没做

C. Tic-tac-toe

 题意：给定一 个一字棋的局面，让你判断是一下列局面中的哪一种

            1、是否合法、

            2、第一个人胜了

            3、第二个人胜了

            4、下一个是第一个人

            5、下一个是第二个人

            6、完了

输入：一个一字棋的局面："0" 代表 第一个人下的棋，"X"代表第二个人下的棋，"."代表空

输出：是哪种局面

题解：很烦的权举，注意我认为最发的是把所有的情况都写出来，再看看读入的局面是否在其中。不过，当时没有而是写了各种if。代码各种长...

代码：

#include <stdio.h>#include <stdlib.h>#include <string.h>const int maxn = 3;char grid[maxn][maxn];int can = 1;int won = 0;int ok1[2] = {0,0};char get(){    char c = getchar();    while(c != '.' && c != '0' && c != 'X'){if(c != '\n' && c != ' '){    can = 0;    printf("illegal\n");    exit(0);}c = getchar();    }    return c;}int line(){    int now = 0;    int i, j, k;    for(i = 0; i < 3; i++){if(grid[i][0] != '.'){    int ok = 1;    for(j = 1; j < 3; j++)if(grid[i][j] != grid[i][0])    ok = 0;    now += ok;    if(ok) {won = grid[i][0];if(grid[i][0] == '0') ok1[0] = 1;else ok1[1] = 1;    }}if(grid[0][i] != '.'){    int ok = 1;    for(j = 1; j < 3; j++)if(grid[j][i] != grid[0][i])    ok = 0;    now += ok;    if(ok){ won = grid[0][i];if(grid[0][i] == '0') ok1[0] = 1;else ok1[1] = 1;    }}    }    if(grid[0][0] == grid[1][1] && grid[1][1] == grid[2][2] && grid[0][0] != '.'){now++;won = grid[0][0];if(grid[1][1] == '0') ok1[0] = 1;else ok1[1] = 1;    }    if(grid[0][2] == grid[1][1] && grid[1][1] == grid[2][0] && grid[0][2] != '.'){now++;won =grid[0][2];if(grid[1][1] == '0') ok1[0] = 1;else ok1[1] = 1;    }    return now;}int main(){    int i, j, n = maxn;    int* ok = ok1;    for(i = 0; i < n; i++)for(j = 0; j < n; j++)    grid[i][j] = get();    int first, second;               first = second = 0;    for(i = 0; i < n; i++)for(j = 0; j < n; j++)    if(grid[i][j] == 'X') first++;    else if(grid[i][j] == '0') second++;    int l = line();    if(can == 0){printf("illegal\n");    }else if(first < second || first-second > 1 || ok[0] == 1 && ok[1] == 1)printf("illegal\n");    else if(l == 1 || first + second == 9 || ok[0] || ok[1])  if(l == 0)    printf("draw\n");else if(ok[1])    if(first > second)printf("the first player won\n");    elseprintf("illegal\n");else {    if(first == second)printf("the second player won\n");    else printf("illegal\n");}    else if(first > second)printf("second\n");    else printf("first\n");    return 0;}

D. Least Cost Bracket Sequence

暂没做