• BZOJ1295 [SCOI2009]最长距离 智商


    大体就是给一个01矩阵,1不可通行,可以破坏k[0,30]个1方格,问最长欧式距离

    脑洞1小时没啥想法。。给正解跪。。居然是跑破坏方块的最短路,然后枚举点对更新答案,只要点对的最短路距离<=k就合法

    naive!

    #include<iostream>
    #include<stdio.h>
    #include<algorithm>
    #include<queue>
    #include<string.h>
    #include<math.h>
    #include<set>
    #include<map>
    #include<vector>
    #include<iomanip>
    #include<stack>
    using namespace std;
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mem(a) memset(a,0,sizeof a)
    #define FOR(a) for(int i=1;i<=a;i++)
    #define sqr(a) (a)*(a)
    
    const int maxn=35*35;
    const ll mod=1e9+7;
    
    int n,m,t;
    
    char buf[maxn];
    int mp[maxn][maxn]; 
    
    struct EDGE{int v,w;};
    vector<EDGE>G[maxn];
    void addedge(int u,int v,int w){
    	G[u].pb((EDGE){v,w});
    }
    struct NODE{
    	int v,d;
    	friend bool operator<(NODE a,NODE b){return a.d>b.d;}
    };
    int d[maxn][maxn],vis[maxn];
    void dij(int from){
    	memset(vis,0,sizeof vis);
    	for(int i=0;i<maxn;i++)d[from][i]=mod;
    	if(!mp[(from+m-1)/m][from%m!=0?from%m:m])d[from][from]=0;
    	else d[from][from]=1;
    	priority_queue<NODE>Q;
    	NODE now=(NODE){from,0};
    	Q.push(now);
    	while(!Q.empty()){
    		NODE now=Q.top();Q.pop();
    		int u=now.v;
    		if(vis[u])continue;
    		vis[u]=1;
    		for(int i=0;i<G[u].size();i++){
    			if(d[from][G[u][i].v]>d[from][u]+G[u][i].w){
    				d[from][G[u][i].v]=d[from][u]+G[u][i].w;
    				NODE nxt=(NODE){G[u][i].v,d[from][G[u][i].v]};
    				Q.push(nxt);
    			}
    		}
    	}
    }
    
    int main(){
    	scanf("%d%d%d",&n,&m,&t);
    	for(int i=1;i<=n;i++){
    		scanf("%s",buf+1);
    		for(int j=1;j<=m;j++)mp[i][j]=buf[j]=='1'?1:0;
    	}
    
    	for(int i=1;i<=n;i++){
    		for(int j=1;j<=m;j++){
    			int x=i,y=j;
    			for(int dx=-1;dx<=1;dx++){
    				for(int dy=-1;dy<=1;dy++){
    					if(dx==dy || dx+dy==0)continue;
    					if(x+dx>n||x+dx<1||y+dy>m||y+dy<1)continue;
    					if(mp[x+dx][y+dy]){
    						addedge((x-1)*m+y,(x+dx-1)*m+y+dy,1);					
    					}else{
    						addedge((x-1)*m+y,(x+dx-1)*m+y+dy,0);
    					}
    				}
    			}	
    		}
    	}
    	for(int i=1;i<=n*m;i++){
    		dij(i);
    	}
    	double ans=0;
    	for(int i=1;i<=n*m;i++){
    		for(int j=1+i;j<=n*m;j++){
    			if(d[i][j]<=t){
    
    				int yy1=i%m==0?m:i%m;
    				int yy2=j%m==0?m:j%m;
    				//if(mp[(i+m-1)/m][yy1])if(d[i][j]==t-1)continue;
    				ans=max(ans,sqrt(sqr((i+m-1)/m-(j+m-1)/m) + sqr(yy1-yy2)));
    			}
    		}
    	}
    	printf("%.6lf
    ",ans);
    }



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  • 原文地址:https://www.cnblogs.com/Drenight/p/8611224.html
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