Google Grad Test Second Round

Problem A Sudoku Checker

总结: 简单模拟题. 时间复杂度 o(n^4)

#include <iostream>

#include <stdio.h>

#include <memory.h>

#include <algorithm>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <deque>

#define MIN(x,y) (x)<(y)?(x):(y)

using namespace std;

 

bool valid;

int matrix[1600][1600];

 

 

void initSet(set<int> &set_it, int n) {

    

    for(int i = 1; i <= n*n; i++)

        set_it.insert(i);

}

 

void smallPart(int x, int y, int n) {

    set<int> tmp;

    initSet(tmp, n);

    

    for(int i = x; i < n; i ++) {

        for(int j = y; j < n; j ++) {

            if(tmp.count(matrix[i][j]) == 0) {

                valid = false;

                return;

            }

            tmp.erase(tmp.find(matrix[i][j]));

        }

    }

}

 

void bigPart(int n) {

    for(int i = 0; i < n*n && valid; i ++) { /* for each line */

        set<int> tmp;

        initSet(tmp, n);

        for(int j = 0; j < n*n && valid; j ++) {

            if(tmp.count(matrix[i][j]) == 0) {

                valid = false;

                return;

            }

            tmp.erase(tmp.find(matrix[i][j]));

        }

    }

 

    for(int j = 0; j < n*n && valid; j ++) {

        set<int> tmp;

        initSet(tmp, n);

        for(int i = 0; i < n*n && valid; i ++) {

            if(tmp.count(matrix[i][j]) == 0) {

                valid = false;

                return;

            }

            tmp.erase(tmp.find(matrix[i][j]));

        }

    }

 

    for(int t = 0; t < n*n; t ++) {

        int i = t / n;

        int j = t % n;

 

        smallPart(i*n, j*n, n);

    }

}

 

int main() {

    freopen("C:\\Users\\vincent\\Dropbox\\workplacce\\joj\\test.txt", "r", stdin);

    

    int T, iCase = 0;

    scanf("%d", &T);

 

    while(T --) {

        valid = true;

        int n;

        scanf("%d", &n);

 

        for(int i = 0; i < n*n; i ++) {

            for(int j = 0; j < n*n; j ++) {

                scanf("%d", &matrix[i][j]);

            }

        }

        bigPart(n);

 

        if(valid) {

            printf("Case #%d: Yes\n", iCase);

        }else {

            printf("Case #%d: No\n", iCase);

        }

    }

 

    return 0;

}

 

Problem B Dragon Maze

思路:

1. 单源点最短路径变形题. 我用到了两个数组分别判断一个点是否已经被加入到队列中, 和是否还能够被更新该点获得的最大价值, 我觉得这是有些低效的, 应当有更优雅的解法

#include <iostream>

#include <stdio.h>

#include <memory.h>

#include <algorithm>

#include <vector>

#include <map>

#include <set>

#include <string>

#include <deque>

#define MIN(x,y) (x)<(y)?(x):(y)

using namespace std;

 

int value[200][200];

bool visited[200][200];

bool added[200][200];

int matrix[200][200];

 

int dir[4][2] = {0,1, 1,0, 0,-1, -1,0}; /* */

const int INF = 0X3F3F3F3F;

 

int sx, sy, ex, ey;

int n, m;

 

 

/* using more space for simplicity */

class Node {

public:

    int x, y, dist;

    Node() {

        x = y = dist = 0;

    }

    Node(int _x, int _y, int _dist):x(_x), y(_y), dist(_dist) {}

};

 

int bfs() {

    deque<Node> queue;

    int minDist = INF;

 

    memset(visited, 0, sizeof(visited));

    memset(value, 0, sizeof(value));

    memset(added, 0, sizeof(added));

 

    queue.push_back(Node(sx, sy, 0));

    added[sx][sy] = 1;

    value[sx][sy] = matrix[sx][sy];

 

    while(!queue.empty()) {

        Node fath = queue.front();

        queue.pop_front();

        visited[fath.x][fath.y] = 1;

        

        if(fath.x == ex && fath.y == ey) { /* give same level nodes opportunity*/

            minDist = fath.dist;

        }

 

        if(fath.dist > minDist) {

            break;

        }

 

        for(int i = 0; i < 4; i ++) {

            int cx = fath.x + dir[i][0];

            int cy = fath.y + dir[i][1];

 

            if(cx < 0 || cy < 0 || cx > n-1 || cy > m-1)

                continue;

 

            if(matrix[cx][cy] < 0) { /* dangerous place */

                continue;

            }

 

            if(!visited[cx][cy]) { /* update */

                value[cx][cy] = max(value[cx][cy], value[fath.x][fath.y]+matrix[cx][cy]);    

                if(!added[cx][cy]) {

                    queue.push_back(Node(cx, cy, fath.dist+1));

                    added[cx][cy] = 1;

                }

            }

        }

    }

 

    return value[ex][ey];

}

 

int main() {

    freopen("C:\\Users\\vincent\\Dropbox\\workplacce\\joj\\test.txt", "r", stdin);

 

    int T, iCase = 0;

    scanf("%d", &T);

    

    while(T --) {

        scanf("%d%d", &n, &m);

        scanf("%d%d%d%d", &sx, &sy, &ex, &ey);

        iCase ++;

 

        for(int i = 0; i < n; i ++) {

            for(int j = 0; j < m; j ++) {

                scanf("%d", &matrix[i][j]);

            }

        }

 

        int res = bfs();

        printf("Case #%d: ", iCase);

        if(res == 0) {

            printf("Mission Impossible\n");

        }else {

            printf("%d\n", res);

        }

    }

 

    return 0;

}

 

Problem E Ignore All My Comments

思路:

1. 每次读取一行, 并且把每次读取的字符放到 string 里, 100kb 装的下

2. 查找符号 '/', 以此为根据向右扩展得到 /*, 向左扩展得到 */, 能使问题简化

3. 我最初用 scanf() 来移位, 结局只能说太惨了, 因为没有 look_back() 功能, //* 这种类型的就判断不出了

 

Problem C Hex

思路

1. 所有不正确的状态: a. 双方棋子相差 1 以上; b. 红方赢, 但蓝方的棋子更多. c. 红方赢, 但红方的棋子路径上没有关键点.

2. 这道题是相当难得, 师兄当初靠抱 ACMer 的大腿才弄过去

 

Problem B Meet and Party

1. 最简单的解法是枚举, 但只能过小数据.

2. 优化, 每次计算一个点到一块区域的长度之和.

3. 最优解法

转载于:https://www.cnblogs.com/zhouzhuo/p/3660768.html