题目大意：给出一张地图，上面有些点有障碍物，现在有T个机会能够移除障碍物，问地图上最长的欧几里得距离是多长。

思路：在原图的基础上建图，f[i]表示的是起点到这里最少需要移除多少个障碍物，然后暴力枚举起点，更新答案即可。

CODE：

#include <map>
#include <queue>
#include <cmath>
#include <cstdio>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 10010
using namespace std;
const int dx[] = {0,1,-1,0,0};
const int dy[] = {0,0,0,1,-1};

map<int,pair<int,int> > G;

int m,n,k;
int head[MAX],total;
int next[MAX],aim[MAX];
bool is_block[MAX];

int f[MAX];
bool v[MAX];

char src[110][110];
int num[110][110],cnt;

inline void Add(int x,int y)
{
	next[++total] = head[x];
	aim[total] = y;
	head[x] = total;
}

inline void SPFA(int start)
{
	static queue<int> q;
	while(!q.empty())	q.pop();
	q.push(start);
	memset(v,false,sizeof(v));
	memset(f,0x3f,sizeof(f));
	f[start] = is_block[start];
	while(!q.empty()) {
		int x = q.front(); q.pop();
		v[x] = false;
		for(int i = head[x]; i; i = next[i])
			if(f[aim[i]] > f[x] + is_block[aim[i]]) {
				f[aim[i]] = f[x] + is_block[aim[i]];
				if(!v[aim[i]]) {
					v[aim[i]] = true;
					q.push(aim[i]);
				}
			}
	}
}

inline double Calc(pair<int,int> p1,pair<int,int> p2)
{
	return sqrt((double)(p1.first - p2.first) * (p1.first - p2.first) + (double)(p1.second - p2.second) * (p1.second - p2.second));
}

int main()
{
	cin >> m >> n >> k;
	for(int i = 1; i <= m; ++i) {
		scanf("%s",src[i] + 1);
		for(int j = 1; j <= n; ++j) {
			num[i][j] = ++cnt;
			G[cnt] = make_pair(i,j);
			if(src[i][j] == '1')
				is_block[cnt] = true;
		}
	}
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			for(int k = 1; k <= 4; ++k) {
				int fx = i + dx[k];
				int fy = j + dy[k];
				if(!num[fx][fy])	continue;
				Add(num[i][j],num[fx][fy]);
				Add(num[fx][fy],num[i][j]);
			}
	double ans = .0;
	for(int i = 1; i <= cnt; ++i) {
		SPFA(i);
		for(int j = 1; j <= cnt; ++j)
			if(f[j] <= k)
				ans = max(ans,Calc(G[i],G[j]));
	}
	cout << fixed << setprecision(6) << ans << endl;
	return 0;
}

BZOJ 1295 SCOI 2009 最长距离 SPFA