这题 是被我自己搞复杂了....
太SB了....
还是porker的关于输出路径的简洁 有效多了
#include <iostream>
#include <cstring>
#include <queue>
#include <stack>
using namespace std;
int ans, n, m;
const int size = 110;
char maze[size][size];
bool vis[size][size];
int arr[10010];
int dir[4][2] = { 1, 0, -1, 0, 0, 1, 0, -1 };
struct data
{
int x, y;
int step;
bool operator<(const data& p)const
{
return step>p.step;
}
}st;
void bfs()
{
data now, next;
int xx, yy;
priority_queue<data> q;
while (!q.empty())
q.pop();
st.x = st.y = st.step = 0;
q.push(st);
vis[0][0] = true;
while (!q.empty())
{
now = q.top();
q.pop();
if (now.x == n - 1 && now.y == m - 1)
{
ans = now.step;
return;
}
for (int i = 0; i < 4; i++)
{
xx = now.x + dir[i][0];
yy = now.y + dir[i][1];
if (xx >= 0 && xx < n && yy >= 0 && yy < m && maze[xx][yy] != ‘X‘ && !vis[xx][yy])
{
vis[xx][yy] = true;
next.x = xx;
next.y = yy;
arr[xx*m + yy] = now.x*m + now.y;
if (maze[xx][yy] == ‘.‘)
next.step = now.step + 1;
else
next.step = now.step + (maze[xx][yy] - ‘0‘) + 1;
q.push(next);
}
}
}
return;
}
int main()
{
cin.sync_with_stdio(false);
stack<int>s;
int num1, x1, y1, num2, x2, y2;
int k, flag;
while (cin >> n >> m)
{
flag = 0;
while (!s.empty())
s.pop();
memset(vis, false, sizeof(vis));
memset(arr, -1, sizeof(arr));
ans = 0;
k = 1;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < m; j++)
{
cin >> maze[i][j];
}
}
bfs();
if (ans)
{
cout << "It takes " << ans << " seconds to reach the target position, let me show you the way." << endl;
for (int i = n*m - 1; i != 0; i = arr[i])
{
s.push(i);
}
int prex = 0, prey = 0;
while( !s.empty() )
{
num1 = s.top();
s.pop();
x1 = num1 / m;
y1 = num1 % m;
cout << k++ << "s:(" << prex << "," << prey << ")->(" << x1 << "," << y1 << ")" << endl;
if (maze[x1][y1] != ‘.‘)
{
for (int i = 0; i < maze[x1][y1] - ‘0‘; i++)
{
cout << k++ << "s:FIGHT AT (" << x1 << "," << y1 << ")" << endl;
}
}
prex = x1;
prey = y1;
}
}
else
{
cout << "God please help our poor hero." << endl;
}
cout << "FINISH" << endl;
}
return 0;
}
这是他帮我修改后过的 贴下起初没过的。。。--纪念下自己的白痴的k ans 做法=-=
1 #include <iostream>
2 #include <cstring>
3 #include <queue>
4 #include <stack>
5 using namespace std;
6
7 int ans , n , m;
8 const int size = 110;
9 char maze[size][size];
10 bool vis[size][size];
11 int arr[10010];
12 int dir[4][2] = {1,0,-1,0,0,1,0,-1};
13
14 struct data
15 {
16 int x , y;
17 int step;
18 bool operator<(const data& p )const
19 {
20 return step>p.step;
21 }
22 }st;
23
24 void bfs( )
25 {
26 data now , next;
27 int xx , yy;
28 priority_queue<data> q;
29 while( !q.empty() )
30 q.pop();
31 st.x = st.y = st.step = 0;
32 q.push( st );
33 vis[0][0] = true;
34 while( !q.empty() )
35 {
36 now = q.top();
37 q.pop();
38 if( now.x == n-1 && now.y == m-1 )
39 {
40 ans = now.step;
41 return;
42 }
43 for( int i = 0 ; i<4 ; i++ )
44 {
45 xx = now.x + dir[i][0];
46 yy = now.y + dir[i][1];
47 if( xx>=0 && xx<n && yy>=0 && yy<m && maze[xx][yy]!=‘X‘ && !vis[xx][yy] )
48 {
49 vis[xx][yy] = true;
50 next.x = xx;
51 next.y = yy;
52 arr[xx*m+yy] = now.x*m+now.y;
53 if( maze[xx][yy] == ‘.‘ )
54 next.step = now.step + 1;
55 else
56 next.step = now.step + (maze[xx][yy] -‘0‘)+1;
57 q.push(next);
58 }
59 }
60 }
61 return;
62 }
63
64 int main()
65 {
66 stack<int>s;
67 int num1 , x1 , y1 , num2 , x2 , y2;
68 int k , flag;
69 while( cin >> n >> m )
70 {
71 flag = 0;
72 while( !s.empty() )
73 s.pop();
74 memset( vis , false , sizeof(vis) );
75 memset( arr , -1 , sizeof(arr) );
76 ans = 0;
77 k = 1;
78 for( int i = 0 ; i<n ; i++ )
79 {
80 for( int j = 0 ; j<m ; j++ )
81 {
82 cin >> maze[i][j];
83 }
84 }
85 bfs( );
86 if( ans )
87 {
88 cout << "It takes " << ans << " seconds to reach the target position, let me show you the way." << endl;
89 for( int i = n*m-1 ; i!=-1 ; i = arr[i] )
90 {
91 s.push( i );
92 }
93 /*
94 while(!s.empty())
95 {
96 cout << s.top() <<endl;
97 s.pop();
98 }
99 */
100 while( k<ans )
101 {
102 num1 = s.top();
103 x1 = num1/m;
104 y1 = num1%m;
105 s.pop();
106 if( maze[x1][y1] ==‘.‘ )
107 {
108 if( !flag )
109 {
110 num2 = s.top();
111 x2 = num2/m;
112 y2 = num2%m;
113 cout<<k<<"s: ("<<x1<<","<<y1<<")->("<<x2<<","<<y2<<")"<<endl;
114 }
115 else
116 {
117 x2 = flag/m;
118 y2 = flag%m;
119 cout<<k<<"s: ("<<x2<<","<<y2<<")->("<<x1<<","<<y1<<")"<<endl;
120 flag = 0;
121 }
122 k++;
123 }
124 else
125 {
126 for( int j = 0 ; j<(maze[x1][y1]-‘0‘) ; j++ )
127 {
128 cout<<k+j<<"s:FIGHT AT ("<<x1<<","<<y1<<")"<<endl;
129 }
130 k+=( maze[x1][y1]-‘0‘ );
131 flag = num1;
132 }
133 }
134 }
135 else
136 {
137 cout << "God please help our poor hero." << endl;
138 }
139 cout << "FINISH" << endl;
140 }
141 return 0;
142 }
洗澡去了~再看部电影 碎觉 =-=
hdu--1026--bfs&&优先队列&&打印路径,布布扣,bubuko.com
时间： 2024-07-31 14:35:25