USACO The Tamworth Two 模拟

一道模拟题不过要担心的是牛或者人在转弯的时候，另一方如果能走，那么要走，不能停留。

还是蛮简单的。

调试输出的话可以看到具体追击过程

Source Code:

/*
ID: wushuai2
PROG: ttwo
LANG: C++
*/
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cmath>
#include <stack>
#include <string>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <vector>
#include <algorithm>
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
#define MOD 1000000007
#define pi acos(-1.0)
#define RV(num) ((num) > 0 ? 0 : 1)

using namespace std;

typedef long long           ll      ;
typedef unsigned long long  ull     ;
typedef unsigned int        uint    ;
typedef unsigned char       uchar   ;

template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;}
template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;}

const double eps = 1e-7      ;
const int M = 660000         ;
const ll P = 10000000097ll   ;
const int INF = 0x3f3f3f3f   ;
const int MAX_N = 20         ;
const int MAXSIZE = 101000000;

ofstream fout ("ttwo.out");
ifstream fin ("ttwo.in");

int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
char gra[11][11];
int dir_f, dir_c;

struct sc{
    int x, y;
}pos[2], nxt_pos[2], sta[2];

bool check(int x, int y){
    if(x >= 1 && x <= 10 && y >= 1 && y <= 10){
        if(gra[x][y] != ‘*‘){
            return true;
        }
    }
    return false;
}

bool finish(){
    if(pos[0].x == pos[1].x && pos[1].y == pos[0].y){
        return true;
    } else  return false;
}

int main(){
    int i, j, k, l, m, n, t, s, c, w, q, u, v, num, val;
    for(i = 1; i <= 10; ++i){
        for(j = 1; j <= 10; ++j){
            fin >> gra[i][j];
            if(gra[i][j] == ‘F‘){
                sta[0].x = i, sta[0].y = j;
                //pos[0].x = i, pos[0].y = j;
            } else if(gra[i][j] == ‘C‘){
                sta[1].x = i, sta[1].y = j;
                //pos[1].x = i, pos[1].y = j;
            }
        }
    }
    int ans = 1;
    int dir_f = 0;
    int dir_c = 0;
    pos[0] = sta[0], pos[1] = sta[1];
    for(;;){
        /*
        for(i = 1; i <= 10; ++i){
            for(j = 1; j <= 10; ++j){
                if(pos[0].x == i && pos[0].y == j){
                    fout << ‘F‘;
                } else if(pos[1].x == i && pos[1].y == j){
                    fout << ‘C‘;
                } else if(gra[i][j] == ‘*‘) fout << ‘*‘;
                else    fout << ‘.‘;
            }
            fout << endl;
        }
        fout << endl;
        */
        nxt_pos[0].x = pos[0].x + dir[dir_f][0];
        nxt_pos[0].y = pos[0].y + dir[dir_f][1];
        nxt_pos[1].x = pos[1].x + dir[dir_c][0];
        nxt_pos[1].y = pos[1].y + dir[dir_c][1];
        if(check(nxt_pos[0].x, nxt_pos[0].y) && check(nxt_pos[1].x, nxt_pos[1].y)){
            pos[0] = nxt_pos[0], pos[1] = nxt_pos[1];
        } else if(!check(nxt_pos[0].x, nxt_pos[0].y) && !check(nxt_pos[1].x, nxt_pos[1].y)){
            dir_f = (dir_f + 1) % 4;
            dir_c = (dir_c + 1) % 4;
        } else if(check(nxt_pos[0].x, nxt_pos[0].y) && !check(nxt_pos[1].x, nxt_pos[1].y)){
            dir_c = (dir_c + 1) % 4;
            pos[0] = nxt_pos[0];
        } else if(!check(nxt_pos[0].x, nxt_pos[0].y) && check(nxt_pos[1].x, nxt_pos[1].y)){
            dir_f = (dir_f + 1) % 4;
            pos[1] = nxt_pos[1];
        }
        if(finish()){
            break;
        }
        ++ans;
        if(ans > 10000){
            ans = 0;
            break;
        }
    }

    fout << ans << endl;
    fin.close();
    fout.close();
    return 0;
}