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  • FZU

    Fire Game 

    题意:

    两个小朋友可以任选一块草地点火,草地可以不同,也可以相同,问最少的烧光草地的时间。

    思路:

    一开始看到这个以为是联通块计数,没想到这道题通过枚举两个起始点作为队列的初始点,每次跑一边bfs即可。

    #include <algorithm>
    #include  <iterator>
    #include  <iostream>
    #include   <cstring>
    #include   <cstdlib>
    #include   <iomanip>
    #include    <bitset>
    #include    <cctype>
    #include    <cstdio>
    #include    <string>
    #include    <vector>
    #include     <cmath>
    #include     <queue>
    #include      <list>
    #include       <map>
    #include       <set>
    using namespace std;
    //#pragma GCC optimize(3)
    //#pragma comment(linker, "/STACK:102400000,102400000")  //c++
    #define lson (l , mid , rt << 1)
    #define rson (mid + 1 , r , rt << 1 | 1)
    #define debug(x) cerr << #x << " = " << x << "
    ";
    #define pb push_back
    #define pq priority_queue
    
    
    
    typedef long long ll;
    typedef unsigned long long ull;
    
    typedef pair<ll ,ll > pll;
    typedef pair<int ,int > pii;
    typedef pair<int,pii> p3;
    
    //priority_queue<int> q;//这是一个大根堆q
    //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
    #define fi first
    #define se second
    //#define endl '
    '
    
    #define OKC ios::sync_with_stdio(false);cin.tie(0)
    #define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
    #define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
    //priority_queue<int ,vector<int>, greater<int> >que;
    
    const ll mos = 0x7FFFFFFFLL;  //2147483647
    const ll nmos = 0x80000000LL;  //-2147483648
    const int inf = 0x3f3f3f3f;
    const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18
    const int mod = 1e9+7;
    
    const double PI=acos(-1.0);
    
    // #define _DEBUG;         //*//
    #ifdef _DEBUG
    freopen("input", "r", stdin);
    // freopen("output.txt", "w", stdout);
    #endif
    /*-----------------------showtime----------------------*/
                const int maxn = 20;
                int n,m;
                string  mp[maxn];
                int dp[maxn][maxn];
                
                int nt[4][4] = {
                    {1,0},{0,1},{-1,0},{0,-1},
                };
                struct node
                {
                    int x,y,step;                
                }q[maxn*maxn];
                int mx,tot;
                void bfs(node a,node b){
                    mx = 0;
                    for(int i=0; i<maxn; i++){
                        for(int j=0; j<maxn; j++)dp[i][j] = inf;
                    }
                    queue<node>que;
                    que.push(a);
                    if(a.x!=b.x||a.y!=b.y)que.push(b);
                    dp[a.x][a.y] = 0;
                    dp[b.x][b.y] = 0;
                    while(!que.empty()){
                        int x = que.front().x,y = que.front().y,s = que.front().step;
                        que.pop();
                        for(int i=0 ; i<4; i++){
                            int nx = x + nt[i][0],ny = y + nt[i][1];
                            if(nx < 0||nx >=n || ny <0||ny>=m)continue;
                            if(mp[nx][ny]=='.')continue;
                            if(dp[nx][ny] > s + 1){
                                    dp[nx][ny] = s + 1;
                                    mx = max(mx,s+1);
                                    node tmp={nx,ny,s+1};
                                    que.push(tmp);
    
                            }
                        }
                    }
                }
                void solve(){
                    cin>>n>>m;
                    
                    for(int i=0; i<n; i++){
                        cin>>mp[i];
                    }
                    int cnt = 0,ans = inf;
                    for(int i=0; i<n ; i++){
                        for(int j=0; j<m; j++){
                          if(mp[i][j]=='#'){
                                q[++cnt].x = i;
                                q[cnt].y = j;
                                q[cnt].step = 0;
                          }
                        }
                    }
                    
                    for(int i=1; i<=cnt; i++)
                    {
    
                        for(int j=i; j<=cnt; j++){
                            bfs(q[i],q[j]);
                            int flag = 1;
                            for(int x = 0; x < n; x++){
                                for(int y = 0; y<m; y++){
                                    if(dp[x][y] >=inf && mp[x][y] == '#')flag= 0;
                                }
                            }
                            if(flag)ans = min(ans, mx);
                            
                        }
                    }
                    if(ans < inf)cout<<ans<<endl;
                    else cout<<-1<<endl;
                }
    int main(){ 
                int T;  cin>>T;
                for(int t=1;t<=T;t++){
                    cout<<"Case "<<t<<": ";
                    solve();
                }
                return 0;
    }
    FZU-2150
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  • 原文地址:https://www.cnblogs.com/ckxkexing/p/9500678.html
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