题目链接：http://codeforces.com/gym/101755/problem/H
题目分析：先bfs一遍怪兽可以到达的点，再bfs人可以走的地方看可不可以到达终点；
很显然读到 2<=n*m<=200000 时，就不可以用二维数组存图了，不过据说因为数据比较水，可以用vector存图；
vector存图AC代码：
1 /* */
2 # include <iostream>
3 # include <stdio.h>
4 # include <string.h>
5 # include <string>
6 # include <cmath>
7 # include <climits>
8 # include <cctype>
9 # include <ctime>
10 # include <algorithm>
11 # include <functional>
12 # include <bitset>
13 # include <set>
14 # include <map>
15 # include <deque>
16 # include <queue>
17 # include <stack>
18 # include <vector>
19 using namespace std;
20
21 const int maxn=2e5+7;
22 vector<char>mp[maxn];
23 vector<int>dis[maxn];
24 vector<int>vis[maxn];
25 int n, m;
26
27 struct node
28 {
29 int x, y;
30 int step;
31 }cur, after;
32
33 int dir[4][2]={{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
34 int sx, sy, fx, fy;
35
36 int check(int x, int y)
37 {
38 if( x>=0 && x<n && y>=0 && y<m )
39 return 1;
40 return 0;
41 }
42
43 void bfs1(int d)
44 {
45 queue<node>q;
46 int i, j;
47 for( i=0; i<n; i++ )
48 {
49 for( j=0; j<m; j++ )
50 {
51 if( mp[i][j]==‘M‘ )
52 {
53 cur.x = i;
54 cur.y = j;
55 cur.step = 0;
56 q.push(cur);
57 dis[i][j] = 1;
58 }
59 }
60 }
61
62 while( !q.empty() )
63 {
64 cur = q.front();
65 q.pop();
66 for(int i=0; i<4; i++ )
67 {
68 after.x = cur.x+dir[i][0];
69 after.y = cur.y+dir[i][1];
70 after.step = cur.step + 1;
71
72 if( check(after.x, after.y) )
73 {
74 if( !dis[after.x][after.y] && after.step<=d )
75 {
76 dis[after.x][after.y] = 1;
77 q.push(after);
78 }
79 }
80 }
81 }
82 }
83
84 void bfs2(int x, int y)
85 {
86 cur.x = x;
87 cur.y = y;
88 cur.step = 0;
89 queue<node>q;
90 q.push(cur);
91 vis[x][y] = 1;
92 if( dis[x][y] )
93 {
94 printf("-1\n");
95 return ;
96 }
97 while( !q.empty() )
98 {
99 cur = q.front();
100 q.pop();
101 if( mp[cur.x][cur.y]==‘F‘ )
102 {
103 printf("%d\n", cur.step);
104 return ;
105 }
106 for(int i=0; i<4; i++ )
107 {
108 after.x = cur.x + dir[i][0];
109 after.y = cur.y + dir[i][1];
110 after.step = cur.step + 1;
111 if( check(after.x, after.y) )
112 {
113 if( !dis[after.x][after.y] && !vis[after.x][after.y])
114 {
115 vis[after.x][after.y] = 1;
116 q.push(after);
117 }
118 }
119 }
120 }
121 printf("-1\n");
122 return ;
123 }
124 int main()
125 {
126 int d, i, j;
127 cin>>n>>m>>d;
128 char s;
129
130 for(i=0; i<n; i++)
131 {
132 mp[i].clear();
133 vis[i].clear();
134 dis[i].clear();
135 }
136
137 for(i=0; i<n; i++ )
138 {
139 for(j=0; j<m; j++ )
140 {
141 cin>>s;
142 mp[i].push_back(s);
143 dis[i].push_back(0);
144 vis[i].push_back(0);
145 }
146 }
147 bfs1(d);
148
149 for(i=0; i<n; i++ )
150 {
151 for( j=0; j<m; j++ )
152 {
153 if( mp[i][j]==‘F‘ )
154 {
155 fx = i;
156 fy = j;
157 }
158 if( mp[i][j]==‘S‘ )
159 {
160 sx = i;
161 sy = j;
162 }
163 }
164 }
165
166 if( dis[fx][fy] )
167 {
168 printf("-1\n");
169 }
170 else
171 {
172 bfs2(sx, sy);
173 }
174 return 0;
175 }
这道题也让我知道了可以用一位数组存图：
详细的见代码注释：
AC代码：
1 /* */
2 # include <iostream>
3 # include <stdio.h>
4 # include <string.h>
5 # include <algorithm>
6 # include <cctype>
7 # include <ctime>
8 # include <functional>
9 # include <cmath>
10 # include <bitset>
11 # include <deque>
12 # include <queue>
13 # include <stack>
14 # include <vector>
15 # include <set>
16 # include <map>
17 # include <climits>
18 using namespace std;
19
20 typedef long long LL;
21 const int maxn=1e6+100;
22 int n, m, t;
23 int a[maxn], d;
24 char str[maxn];
25 bool vis[maxn];///标记是否已经访问过
26 int dis[maxn];///记录步数
27 int dd[maxn];///dd[]为0,说明怪兽到不了,dd不为0说明怪兽可以到此处
28 int cnt, fx, fy, sx, sy;
29 int dir[4][2]={{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
30 struct node
31 {
32 int x;
33 int y;
34 }g[maxn], cur, after;
35
36 bool check(int a, int b)
37 {
38 if( a>=1 && b>=1 && a<=n && b<=m && !dd[a*m+b] )
39 return true;
40 return false;
41 }
42
43 void bfs()
44 {
45 queue<node>q;
46 cur.x = sx;
47 cur.y = sy;
48 vis[cur.x*m+cur.y] = 1;
49 q.push(cur);
50
51 while( !q.empty())
52 {
53 cur = q.front();
54 q.pop();
55 for(int i=0; i<4; i++ )
56 {
57 int xx = cur.x + dir[i][0];
58 int yy = cur.y + dir[i][1];
59
60 if( check(xx, yy) && !vis[xx*m+yy])
61 {
62 dis[xx*m+yy] = dis[cur.x*m+cur.y]+1;
63 vis[xx*m+yy] = 1;
64 after.x = xx;
65 after.y = yy;
66 q.push(after);
67 }
68 }
69 }
70 return ;
71 }
72
73 void bfss()///广搜怪兽可以到达的地方
74 {
75 queue<node>q;
76 for(int i=0; i<cnt; i++ )
77 q.push(g[i]);
78
79 while( !q.empty() )
80 {
81 cur=q.front();
82 q.pop();
83
84 if( dd[cur.x*m+cur.y]==0 )
85 continue;
86
87 for(int i=0; i<4; i++ )
88 {
89 int xx = cur.x + dir[i][0];
90 int yy = cur.y + dir[i][1];
91 if( check(xx, yy) )
92 {
93 after.x = xx;
94 after.y = yy;
95 dd[xx*m+yy] = dd[cur.x*m+cur.y]-1;
96 q.push(after);
97 }
98 }
99 }
100 return ;
101 }
102
103 int main()
104 {
105 while( ~ scanf("%d %d %d", &n, &m, &d) )
106 {
107 getchar();
108 for(int i=1; i<=n; i++ )
109 scanf("%s", &str[i*m+1]);///一维数组存图
110
111 memset(vis, false, sizeof(vis));
112 memset(dd, 0, sizeof(dd));
113 memset(dis, 0, sizeof(dis));
114
115 for(int i=1; i<=n; i++ )
116 {
117 for(int j=1; j<=m; j++ )
118 {
119 if( str[i*m+j]==‘S‘ )
120 {
121 sx = i;
122 sy = j;
123 }
124
125 else if( str[i*m+j]==‘F‘ )
126 {
127 fx = i;
128 fy = j;
129 }
130
131 else if( str[i*m+j]==‘M‘ )
132 {
133 dd[i*m+j] = d+1;
134 cur.x = i;
135 cur.y = j;
136 g[cnt++] = cur;
137 }
138 }
139 }
140
141 bfss();
142 if( dd[sx*m+sy] || dd[fx*m+fy] )
143 {
144 printf("-1\n");
145 }
146 else
147 {
148 bfs();
149 if( !vis[fx*m+fy] )
150 printf("-1\n");
151 else
152 printf("%d\n", dis[fx*m+fy]);
153 }
154 }
155 return 0;
156 }
原文地址：https://www.cnblogs.com/wsy107316/p/11327168.html
时间： 2024-10-10 17:13:00