FZU 2150 枚举＋BFS

Posted 2020-08-14 EqTStar

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 }