爬格子呀6-4、6-5、6-9

今天又写了三个,准确的说是昨天写的,但是昨晚很久才睡的,所以就把它放到今天发了,主题是bfs,学会了还是很开心的;
代码如下:
6-4:

#include<cstdio>
#include<iostream>
#include<queue>

using namespace std;
int a[8][8];
int b1, b2, d1, d2;
const int INF = 0x3f3f3f;
int dir[8][2] = { {1,2},{1,-2},{-1,2},{-1,-2},{2,1},{-2,1},{-2,-1},{2,-1} };

bool legal(int x, int y) {
    return x >= 0 && x < 8 && y >= 0 && y < 8;
}

struct poi {
    int x, y;
    poi():x(0),y(0){}
};
poi po1, po2;

//整个bfs没有涉及到递归调用
//因为每个点的相关点在同一层中被遍历到的时候路径都是一样的,所以就不存在记录距离大小进行比较的问题
int bfs() {
    queue<poi>p;
    p.push(po1);//放进第一个点
    a[po1.x][po1.y] = 0;//标记
    while (p.size()) {//这个条件的确立就是保证他可以遍历所有的点
        poi mid = p.front(), mid2;//递归的开始,也不能算是递归
        p.pop();
        int x = mid.x, y = mid.y;
        if (x == po2.x&&y == po2.y)
            return a[x][y];//这个是判断条件
        for (int i = 0; i < 8; i++) {
            mid2.x = x + dir[i][0];
            mid2.y = y + dir[i][1];
            if (legal(mid2.x, mid2.y) && a[mid2.x][mid2.y] == INF)
                p.push(mid2);//放入相关点,或者连接点
            a[mid2.x][mid2.y] = a[x][y] + 1;//这个是记录距离用的
        }
    }
    return INF;
}

int main() {
    char c1, c2;
    cin >> c1 >> b1 >> c2 >> b2;
    po1.x = 8 - b1;
    po2.x = 8 - b2;
    po1.y = c1 - 'a';
    po2.y = c2 - 'a';

    for (int i = 0; i < 8; i++)
        for (int j = 0; j < 8; j++)
            a[i][j] = INF;
    cout << bfs();
    return 0;

}

6-5:

#include<cstdio>
#include<iostream>
#include<queue>
using namespace std;
int a[20][20], m, n, k, b[20][20];
int dir[4][2] = { {1,0},{-1,0},{0,1},{0,-1} };
const int INF = 0x3f3f3f3f;

bool legal(int x, int y) {
    return x >= 0 && x < m&&y >= 0 && y < n;
}

struct poi {
    int x, y, t;
    poi():x(0),y(0),t(0){}
};
poi po1, po2;

//a为最短路径值,b为血量分布;
int bfs() {
    queue<poi>p;
    p.push(po1);
    a[0][0] = 0;
    while (p.size()) {
        poi mid = p.front(), mid1;
        p.pop();
        if (mid.x == po2.x&&mid.y == po2.y)
            return a[mid.x][mid.y];
        for (int i = 0; i < 4; i++) {
            mid1.x = mid.x + dir[i][0];
            mid1.y = mid.y + dir[i][1];
            if (legal(mid1.x, mid1.y))
            {
                mid1.t = mid.t + b[mid1.x][mid1.y];
                if (mid1.t < k) { 
                    a[mid1.x][mid1.y] = a[mid.x][mid.y] + 1;
                    p.push(mid1);
                }
            }
        }
    }
    return INF;
}

int main() {
    cin >> m >> n >> k;
    int i, j;
    for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++)
        {
            cin >> b[i][j];
            a[i][j] = INF;
        }
    }
    po2.x = m - 1;
    po2.y = n - 1;
    cout << bfs();
    return 0;
}

6-9:

#include<cstdio>
#include<iostream>
#include<stack>
#include<list>
#include<map>
#include<string>
using namespace std;

struct brand {
    string suit;
    string rank;
};
list<stack<brand>>a;

bool can_move(brand x,brand y) {
    return x.suit == y.suit || x.rank == y.rank;
}

void move(list<stack<brand>>::iterator &it, list<stack<brand>>::iterator &ittt) {
    brand k;
    k = it->top();
    it->pop();
    ittt->push(k);
    if (it->empty())
        a.erase(it);
    it = ittt;
}

bool a_move(list<stack<brand>>::iterator &it) {
    list<stack<brand>>::iterator itt, ittt;
    itt = it;
    advance(itt, -1);
    ittt = itt;
    if (distance(itt, a.begin()) >= 2)
        advance(ittt, -2);
    bool can1, can3;
    can1 = can_move(it->top(), itt->top());
    can3 = can_move(it->top(), ittt->top());
    if (can1&&can3)
    {
        move(it, ittt);
        return true;
    }
    else if (can3)
    {
        move(it, ittt);
        return true;
    }
    else if (can1)
    {
        move(it, itt);
        return true;
    }
    else
        return false;
}

int main() {
    int i = 52;
    brand mid;
    stack<brand>mid1;
    string ini;
    while (i--) {
        cin >> ini;
        mid.rank = ini.front();
        mid.suit = ini.substr(1);
        mid1.push(mid);
        a.push_back(mid1);
    }
    list<stack<brand>>::iterator it=a.begin(), itt, ittt;
    it++;
    while (it != a.end()) {
        while (a_move(it)) {}
        it++;
    }
    cout << a.size() << "piles remaining:";
    for (auto i : a) 
        cout << i.size() << ' ';
    return 0;
}
原文地址:https://www.cnblogs.com/romaLzhih/p/9489854.html