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  • 【习题 6-5 UVA-1600】Patrol Robot

    【链接】 我是链接,点我呀:)
    【题意】

    在这里输入题意

    【题解】

    设dis[x][y][z]表示到(x,y)连续走了z个墙的最短路 bfs一下就ok

    【代码】

    /*
      	1.Shoud it use long long ?
      	2.Have you ever test several sample(at least therr) yourself?
      	3.Can you promise that the solution is right? At least,the main ideal
      	4.use the puts("") or putchar() or printf and such things?
      	5.init the used array or any value?
      	6.use error MAX_VALUE?
      	7.use scanf instead of cin/cout?
    */
    #include <bits/stdc++.h>
    using namespace std;
    
    const int N = 20;
    
    int dx[4] = {0,0,1,-1};
    int dy[4] = {1,-1,0,0};
    
    int k,n,m;
    int a[N+10][N+10],dis[N+10][N+10][N+10];
    queue <pair<int,pair<int,int> > > dl;
    
    int main(){
    	#ifdef LOCAL_DEFINE
    	    freopen("F:\c++source\rush_in.txt", "r", stdin);
    	#endif
    	ios::sync_with_stdio(0),cin.tie(0);
    	int T;
    	cin >> T;
    	while (T--){
    		cin >> n >> m;
    		cin >> k;
    		for (int i = 1;i <= n;i++)
    			for (int j = 1;j <= m;j++)
    				cin >> a[i][j];
    		memset(dis,255,sizeof dis);
    		dis[1][1][0] = 0;
    		dl.push(make_pair(1,make_pair(1,0)));
    		while (!dl.empty()){
    			pair <int,pair<int,int> > temp = dl.front();
    			int x = temp.first,y = temp.second.first,num = temp.second.second;
    			dl.pop();
    			for (int i = 0;i < 4;i++){
    			 	int tx = x+dx[i],ty = y + dy[i];
    			 	if (tx >=1 && tx <= n && ty >=1 && ty <= m){
    			 	 	if (a[tx][ty]){
    			 	 	 	int num1 = num+1;
    			 	 	 	if (num1 <= k){
    			 	 	 	 	if (dis[tx][ty][num1]==-1){
    			 	 	 	 	 	dis[tx][ty][num1] = dis[x][y][num] + 1;
    			 	 	 	 	 	dl.push(make_pair(tx,make_pair(ty,num1)));
    			 	 	 		}
    			 	 	 	}
    			 	 	}else{
    			 	 	 	int num1 = 0;
    			 	 	 	if (dis[tx][ty][num1]==-1){
    						 	dis[tx][ty][num1] = dis[x][y][num] + 1;
    			 	 	 	 	dl.push(make_pair(tx,make_pair(ty,num1)));			 	 	 	 	
    			 	 	 	}
    			 	 	}
    			 	}
    			}
    		}
    		int ans = -1;
    		for (int i = 0;i <= k;i++)
    			if (dis[n][m][i]!=-1){
    				if (ans==-1){
    				 	ans = dis[n][m][i];
    				}else{
    				 	ans = min(ans,dis[n][m][i]);
    				}
    			}
    		cout << ans << endl;
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7875307.html
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