【BZOJ 1295】 [SCOI2009] The Longest Distance

【Link】 I am the link, click on me :)【Title】

enter question here

【answer】

Change your thinking.
We find the number of obstacles that reach the least traversed between any two points.

Finally, it is good to find the enumeration answer with the number of obstacles less than or equal to T.

(The minimum obstacle between any two points can be obtained by a method similar to spfa
(a very clever conversion

【Code】

#include <bits/stdc++.h>
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define all(x) x.begin(),x.end()
#define pb push_back
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;

const double pi = acos(-1);
const int dx[4] = {0,0,1,-1};
const int dy[4] = {1,-1,0,0};
const int N = 30;

int n,m,t,dis[N*N+10][N*N+10];;
vector<pair<int,int> > g[N*N+10];
char s[N+10][N+10];
queue<int> dl;
bool inq[N*N+10];

void deal_with(int x,int y){
    memset(inq,0,sizeof inq);
    int S = (x-1)*m+y;
    memset(dis[S],255,sizeof dis[S]);
    dis[S][S] = s[x][y]-'0';
    dl.push(S);
    inq[S] = true;
    while (!dl.empty()){
        int x = dl.front();dl.pop();
        inq[x] = false;
        for (int i = 0;i < (int)g[x].size();i++){
            int y = g[x][i].first,cost = g[x][i].second;
            if (dis[S][y]==-1 || dis[S][y]>dis[S][x]+cost) {
                dis[S][y] = dis[S][x]+cost;
                if (!inq[y]){
                    inq[y] = true;
                    dl.push(y);
                }
            }

        }
    }
}

double sqr(double x){
    return x*x;
}

int main(){
    #ifdef LOCAL_DEFINE
        freopen("rush_in.txt", "r", stdin);
    #endif
    ios::sync_with_stdio(0),cin.tie(0);
    cin >> n >> m >> t;
    rep1(i,1,n) cin >> (s[i]+1);
    rep1(i,1,n)
        rep1(j,1,m)
            rep1(k,0,3){
                int tx = i + dx[k],ty = j+dy[k];
                if (tx>=1 && tx<= n && ty>=1 && ty<=m){
                    int x = (i-1)*m+j,y = (tx-1)*m+ty;
                    g[x].push_back(make_pair(y,s[tx][ty]-'0'));
                }
            }
    rep1(i,1,n)
        rep1(j,1,m)
            deal_with(i,j);

    double ans = 0;
    rep1(i,1,n)
        rep1(j,1,m)
            rep1(ii,1,n)
                rep1(jj,1,m){
                    int x = (i-1)*m+j,y = (ii-1)*m+jj;
                    if (dis[x][y]<=t){
                        ans = max(ans,sqrt(sqr(ii-i)+sqr(jj-j)));
                    }
                }
    cout<<fixed<<setprecision(6)<<ans<<endl;
    return 0;
}