2022春季 哈工大 硕士算法设计与分析 实验二 搜索算法
作者：互联网
单向A*：
v>
算法思想：每次都选择权值最小的结点拓展并记录路径，
如果选择结点为终点，
则找到一条代价最小路径
算法使用优先队列实现
#include <iostream>
#include <cstring>
#include <cmath>
#include <utility>
#include <map>
#include <queue>
using namespace std;
const int maxn = 1010;
string graph[maxn];
int cost[maxn][maxn];
bool vis[maxn][maxn];
int n, m, sx, sy, tx, ty;
// 8个方向
int dir[8][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1}};
struct node
{
int x, y;
double dis;
struct node* pre;
node(int x, int y, double dis, node *pre)
{
this->x = x;
this->y = y;
this->dis = dis;
this->pre = pre;
};
};
map<pair<int, int>, node*> hashmap;
int manha(int x, int y)
{
return abs(x - tx) + abs(y - ty);
}
class Compare_Node_Pointer
{
public:
bool operator () (node* &a, node* &b) const
{
return a->dis + manha(a->x, a->y) > b->dis + manha(b->x, b->y);
}
};
// 记录最后的相遇点，与最短距离
node *ltemp = NULL, *rtemp = NULL;
double maxDis = 0x7fffffff;
void Astar(node* &s)
{
memset(vis, 0, sizeof(vis));
// 正向队列和反向队列
priority_queue<node*, vector<node*>, Compare_Node_Pointer> Q;
Q.push(s);
// dis[u->x][u->y] = 0;
while(!Q.empty())
{
node* p;
p = Q.top();
Q.pop();
if(p->x == tx && p->y == ty)
{
if(ltemp == NULL)
ltemp = p;
// else
// if(ltemp->dis > p->dis)
// ltemp = p;
return;
}
if(vis[p->x][p->y]) continue;
vis[p->x][p->y] = 1;
for(int i = 0; i < 8; i++)
{
int vx = p->x + dir[i][0];
int vy = p->y + dir[i][1];
double w = i < 4 ? 1.0 : (double)sqrt(2.0);
if(vx < 0 || vy < 0 || vx >= n || vy >= m || graph[vx][vy] == '#') continue;
node* v;
if(hashmap[pair<int, int> (vx, vy)] != NULL)
{
v = hashmap[pair<int, int> (vx, vy)];
}
else
{
v = new node(vx, vy, 0x7fffffff, NULL);
hashmap[pair<int, int> (vx, vy)] = v;
}
if(v->dis > p->dis + w + cost[vx][vy])
{
v->dis = p->dis + w + cost[vx][vy];
v->pre = p;
Q.push(v);
}
}
}
}
int main()
{
memset(cost, 0, sizeof(cost));
cout << "Please input row of the map" << endl;
cout << "空地用.表示 沙漠用y表示 溪流用r表示 障碍用#表示" << endl;
cin >> n;
for(int i = 0; i < n; i++)
{
cin >> graph[i];
for(int j = 0; j < graph[i].size(); j++)
{
if(graph[i][j] == 'y')
cost[i][j] = 4;
else if(graph[i][j] == 'r')
cost[i][j] = 2;
}
}
m = graph[0].size();
cout << "Please input start and end points" << endl;
cin >> sx >> sy;
cin >> tx >> ty;
cost[sx][sy] = 0;
node* s = new node(sx, sy, 0, NULL);
hashmap[pair<int, int> (sx, sy)] = s;
Astar(s);
if(ltemp != NULL)
{
cout << "DistCost: " << ltemp->dis << endl;
node *q = ltemp;
while(q)
{
if(q == s)
graph[q->x][q->y] = 'S';
else if(q->x == tx && q->y == ty)
graph[q->x][q->y] = 'T';
else
graph[q->x][q->y] = '&';
q = q->pre;
}
for(int i = 0; i < n; i++)
cout << graph[i] << endl;
}
return 0;
}
双向A*：
v>
算法思想：分别从起点和终点使用A*算法，
每次选择两个队列中拥有最小权值的结点进行拓展，
并记录每个点是由来自哪个开始点拓展的，
如果相遇则记录相遇时两端路程加和的最小值。
算法使用两个优先队列实现
#include <iostream>
#include <cstring>
#include <cmath>
#include <utility>
#include <map>
#include <queue>
using namespace std;
const int maxn = 1010;
string graph[maxn];
int cost[maxn][maxn];
bool vis[maxn][maxn];
int subj[maxn][maxn];
int n, m, sx, sy, tx, ty;
// 8个方向
int dir[8][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}, {1, 1}, {-1, -1}, {1, -1}, {-1, 1}};
struct node
{
int x, y;
double dis;
struct node* pre;
node(int x, int y, double dis, node *pre)
{
this->x = x;
this->y = y;
this->dis = dis;
this->pre = pre;
};
};
map<pair<int, int>, node*> hashmap;
int manha(int x, int y, int Tx, int Ty)
{
return abs(x - Tx) + abs(y - Ty);
}
class Compare_Node_Pointer
{
public:
bool operator () (node* &a, node* &b) const
{
if(subj[a->x][a->y] == 0)
return a->dis + manha(a->x, a->y, tx, ty) > b->dis + manha(b->x, b->y, tx, ty);
else
return a->dis + manha(a->x, a->y, sx, sy) > b->dis + manha(b->x, b->y, sx, sy);
}
};
// 记录最后的相遇点，与最短距离
node *ltemp = NULL, *rtemp = NULL;
double maxDis = 0x7fffffff;
void Astar(node* &s, node* &t)
{
memset(vis, 0, sizeof(vis));
memset(subj, -1, sizeof(subj));
// 正向队列和反向队列
priority_queue<node*, vector<node*>, Compare_Node_Pointer> Q;
priority_queue<node*, vector<node*>, Compare_Node_Pointer> Q1;
Q.push(s);
Q1.push(t);
subj[s->x][s->y] = 0;
subj[t->x][t->y] = 1;
// dis[u->x][u->y] = 0;
while(!Q.empty() || !Q1.empty())
{
node* p;
if(Q.empty() || (!Q1.empty() && Q.top() > Q1.top()))
{
p = Q1.top();
Q1.pop();
}
else
{
p = Q.top();
Q.pop();
}
if(vis[p->x][p->y]) continue;
vis[p->x][p->y] = 1;
for(int i = 0; i < 8; i++)
{
int vx = p->x + dir[i][0];
int vy = p->y + dir[i][1];
double w = i < 4 ? 1.0 : (double)sqrt(2.0);
if(vx < 0 || vy < 0 || vx >= n || vy >= m || graph[vx][vy] == '#') continue;
node* v;
if(hashmap[pair<int, int> (vx, vy)] != NULL)
{
v = hashmap[pair<int, int> (vx, vy)];
}
else
{
v = new node(vx, vy, 0x7fffffff, NULL);
hashmap[pair<int, int> (vx, vy)] = v;
}
// 判断是否相遇，若相遇，则记录总距离最小的相遇点
// 相遇时并不将v放入队列
if(subj[p->x][p->y] == 0 && subj[v->x][v->y] == 1)
{
double c = i < 4 ? 1 : sqrt(2);
if(maxDis > p->dis + v->dis + c + cost[tx][ty])
{
maxDis = p->dis + v->dis + c + cost[tx][ty];
ltemp = p, rtemp = v;
}
continue;
}
else if(subj[p->x][p->y] == 1 && subj[v->x][v->y] == 0)
{
double c = i < 4 ? 1 : sqrt(2);
if(maxDis > p->dis + v->dis + c + cost[tx][ty])
{
maxDis = p->dis + v->dis + c + cost[tx][ty];
ltemp = v, rtemp = p;
}
continue;
}
if(v->dis > p->dis + w + cost[vx][vy])
{
v->dis = p->dis + w + cost[vx][vy];
if(subj[p->x][p->y] == 0)
{
v->pre = p;
subj[v->x][v->y] = 0;
Q.push(v);
}
else
{
v->pre = p;
subj[v->x][v->y] = 1;
Q1.push(v);
}
}
}
}
}
int main()
{
memset(cost, 0, sizeof(cost));
cout << "Please input row of the map" << endl;
cout << "空地用.表示 沙漠用y表示 溪流用r表示 障碍用#表示" << endl;
cin >> n;
for(int i = 0; i < n; i++)
{
cin >> graph[i];
for(int j = 0; j < graph[i].size(); j++)
{
if(graph[i][j] == 'y')
cost[i][j] = 4;
else if(graph[i][j] == 'r')
cost[i][j] = 2;
}
}
m = graph[0].size();
cout << "Please input start and end points" << endl;
cin >> sx >> sy;
cin >> tx >> ty;
cost[sx][sy] = 0;
node* s = new node(sx, sy, 0, NULL);
node* t = new node(tx, ty, 0, NULL);
hashmap[pair<int, int> (sx, sy)] = s;
hashmap[pair<int, int> (tx, ty)] = t;
Astar(s, t);
if(ltemp != NULL)
{
cout << "DistCost: " << maxDis << endl;
node *q = ltemp;
while(q)
{
if(q == s)
graph[q->x][q->y] = 'S';
else
graph[q->x][q->y] = '&';
q = q->pre;
}
q = rtemp;
while(q != t)
{
graph[q->x][q->y] = '%';
q = q->pre;
}
graph[q->x][q->y] = 'T';
for(int i = 0; i < n; i++)
cout << graph[i] << endl;
}
return 0;
}
标签：node,cost,int,搜索算法,哈工大,2022,vx,vy,dis 来源： https://www.cnblogs.com/WTSRUVF/p/16350514.html