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;
}