FZU 2150 枚举＋BFS
题意
给个N＊M的田地，有些地方是草地有些是空地，现在有两个小孩，开始放火。放火只能在草地上进行，并且第一块点火的草地烧掉不需要时间，且每一块着火的草地，每一秒都会点着烧掉周围四块草地。问两个小孩最少要多少时间能把这片田地上的草烧完。每个小孩只有一次放火机会。
分析
N ， M 不会很大，直接枚举两个人点的草地位置，分别bfs，在所有解里取最小值即可。
代码1
/* When all else is lost the future still remains. */
/* You can be the greatest */
#define rep(X,Y,Z) for(int X=(Y);X<(Z);X++)
#define drep(X,Y,Z) for(int X=(Y);X>=(Z);X--)
#define fi first
#define se second
#define mk(X,Y) make_pair((X),(Y))
#define inf 0x3f3f3f3f
#define clr(X,Y) memset(X,Y,sizeof(X))
#define pb push_back
//head
#include <iostream>
#include <stdio.h>
#include <queue>
#include <algorithm>
#include <string>
#include <map>
#include <string.h>
using namespace std;
#define maxN 15
#define maxM 15
int dx[] = {-1,0,1,0};
int dy[] = {0,-1,0,1};
int val[2][maxN][maxM];
int mark[2][maxN][maxM];
bool mp[maxN][maxM];
int n , m;
int check_val(int,int);
int check_val(int n , int m){
int ans = 0;
rep(i,1,n+1) rep(j,1,m+1){
if(mp[i][j]){
ans = max(ans,min(val[1][i][j],val[0][i][j]));
}
}
return ans;
}
int dfs(int x , int y , int cnt){
queue<pair<int,int> > Q;
val[cnt][x][y] = 0;
mark[cnt][x][y] = 1;
Q.push(mk(x,y));
while(!Q.empty()){
int nx = Q.front().fi;
int ny = Q.front().se;
Q.pop();
rep(i,0,4){
int tx = nx + dx[i];
int ty = ny + dy[i];
if(!mp[tx][ty]) continue;
if(mark[cnt][tx][ty]) continue;
val[cnt][tx][ty] = val[cnt][nx][ny] + 1;
mark[cnt][tx][ty] = 1;
Q.push(mk(tx,ty));
}
}
if(cnt == 1) return check_val(n,m);
int ans = inf;
rep(i,x,n+1){
int t = (i == x) ? (y) : 1;
rep(j,t,m+1){
if(mp[i][j]){
rep(k,1,n+1) rep(l,1,m+1) mark[1][k][l] = 0;
rep(k,1,n+1) rep(l,1,m+1) val[1][k][l] = inf;
ans = min(ans,dfs(i,j,1));
}
}
}
return ans;
}
int main(){
int T;
while(~scanf("%d",&T)) rep(ca,1,T+1){
clr(mp,0);
scanf("%d %d\n",&n,&m);
char t;
rep(i,1,n+1) {
rep(j,1,m+1){
scanf("%c",&t);
mp[i][j] = (t == '.') ? 0 : 1;
}
getchar();
}
int ans = inf;
rep(i,1,n+1) rep(j,1,m+1){
if(mp[i][j]){
clr(val,inf);
clr(mark,0);
ans = min(ans,dfs(i,j,0));
}
}
printf("Case %d: %d\n",ca,(ans>=inf)?-1:ans);
}
return 0;
}
代码2
1 /* When all else is lost the future still remains. */
2 /* You can be the greatest */
3 #define rep(X,Y,Z) for(int X=(Y);X<(Z);X++)
4 #define drep(X,Y,Z) for(int X=(Y);X>=(Z);X--)
5 #define fi first
6 #define se second
7 #define mk(X,Y) make_pair((X),(Y))
8 #define inf 0x3f3f3f3f
9 #define clr(X,Y) memset(X,Y,sizeof(X))
10 #define pb push_back
11 //head
12 #include <iostream>
13 #include <stdio.h>
14 #include <queue>
15 #include <algorithm>
16 #include <string>
17 #include <map>
18 #include <string.h>
19 using namespace std;
20 #define maxN 20
21 #define maxM 20
22 int dx[] = {-1,0,1,0};
23 int dy[] = {0,-1,0,1};
24 int n , m;
25 bool mp[maxN][maxM];
26 int val[2][maxN][maxM];
27 bool mark[2][maxN][maxM];
28 void init(int cnt){
29 rep(i,1,n+1) rep(j,1,m+1){
30 val[cnt][i][j] = inf;
31 mark[cnt][i][j] = 0;
32 }
33 return ;
34 }
35 int check_val(){
36 int ans = 0;
37 rep(i,1,n+1) rep(j,1,m+1){
38 if(mp[i][j]) ans = max(ans,min(val[0][i][j],val[1][i][j]));
39 }
40 return ans;
41 }
42 void bfs(int x , int y , int cnt){
43 queue<pair<int,int> > Q;
44 Q.push(mk(x,y));
45 mark[cnt][x][y] = 1;
46 val[cnt][x][y] = 0;
47 while(!Q.empty()){
48 int nx = Q.front().fi;
49 int ny = Q.front().se;
50 Q.pop();
51 rep(i,0,4){
52 int tx = nx + dx[i];
53 int ty = ny + dy[i];
54 if(!mp[tx][ty]) continue;
55 if(mark[cnt][tx][ty]) continue;
56 val[cnt][tx][ty] = val[cnt][nx][ny] + 1;
57 mark[cnt][tx][ty] = 1;
58 Q.push(mk(tx,ty));
59 }
60 }
61 }
62 int main(){
63
64 int T;
65 scanf("%d",&T);
66 rep(ca,1,T+1){
67 int ans = inf;
68 clr(mp,0);
69 scanf("%d %d\n",&n,&m);
70 char t;
71 rep(i,1,n+1) {
72 rep(j,1,m+1){
73 scanf("%c",&t);
74 mp[i][j] = (t == '.') ? 0 : 1;
75 }
76 getchar();
77 }
78 rep(rx,1,n+1) rep(ry,1,m+1){
79 if(!mp[rx][ry]) continue;
80 init(0); init(1);
81 bfs(rx,ry,0);
82 rep(ax,rx,n+1){
83 int t = (ax == rx) ? ry : 1;
84 rep(ay,t,m+1){
85 if(!mp[ax][ay]) continue;
86 init(1);
87 bfs(ax,ay,1);
88 ans = min(ans,check_val());
89 }
90 }
91 }
92 printf("Case %d: %d\n",ca,(ans>=inf)?-1:ans);
93
94 }
95 return 0;
96 }
