[luogu4162 SCOI2009] 最长距离（最短路）

传送门

Solution

题目是最长路，其实是最短路ヽ(ー_ー)ノ
把进入障碍点的边设为1，其他为0。枚举每个点为起点找距离<=T的点，更新答案

Code

//By Menteur_Hxy
#include <queue>
#include <cmath>
#include <cstdio>
#include <vector>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define Re register 
#define Ms(a,b) memset(a,(b),sizeof(a))
#define Fo(i,a,b) for(Re int i=(a),_=(b);i<=_;i++)
#define Ro(i,a,b) for(Re int i=(b),_=(a);i>=_;i--)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;

inline LL read() {
	LL x=0,f=1;char c=getchar();
	while(!isdigit(c)) {if(c=='-')f=-f;c=getchar();}
	while(isdigit(c)) x=(x<<1)+(x<<3)+c-48,c=getchar();
	return x*f;
}

const int N=31;
int n,m,t;
double ans;
bool vis[N*N];
int mp[N*N],dis[N*N][N*N];
char s[N];
vector <int> V[N*N];

inline int id(int x,int y) {return x*m+y-m;}
inline void add(int x,int y) {V[x].push_back(y);}

priority_queue <PII,vector<PII>,greater<PII> >Q;
void Dij(int S,int *d) {
	Ms(vis,0);
	Q.push(PII(d[S]=0,S));
	while(!Q.empty()) {
		int u=Q.top().second,v; Q.pop();
		if(vis[u]) continue; vis[u]=1;
		int siz=V[u].size();
		Fo(i,0,siz-1) if(d[(v=V[u][i])]>d[u]+mp[v]) 
			Q.push(PII(d[v]=d[u]+mp[v],v));
	}
}

int main() {
	n=read(),m=read(),t=read();
	Fo(i,1,n) {
		scanf("%s",s+1);
		Fo(j,1,m) mp[id(i,j)]=s[j]-'0';
	}
	Fo(i,1,n) Fo(j,1,m) {
            if(i>1) add(id(i-1,j),id(i,j));
            if(j>1) add(id(i,j-1),id(i,j));
            if(i<n) add(id(i+1,j),id(i,j));
            if(j<m) add(id(i,j+1),id(i,j));
        }
	Ms(dis,0x3f);
	Fo(i,1,n) Fo(j,1,m) Dij(id(i,j),dis[id(i,j)]);
	Fo(x1,1,n) Fo(y1,1,m) if(!mp[id(x1,y1)])
		Fo(x2,1,n) Fo(y2,1,m) if(dis[id(x1,y1)][id(x2,y2)]<=t)
			ans=max(ans,(double)sqrt((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)));
	printf("%.6lf",ans);
	return 0;
}