BZOJ 1295 最长距离(最短路)

考虑到矩阵尺寸的关系，可以枚举起点和终点，并且判断是否可行即可。

判断起点和终点是否可以通过挖空至多T个障碍联通。实际上就是求起点到终点的最短路。

所以我们先建好图，然后求以每个方格为起始点的最短路，复杂度O(n*m^2*logm).

枚举起点和终点更新答案的复杂度是O(n^2*m^2).

总复杂度就是(nm(nm+mlogm)).

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi 3.1415926535
# define eps 1e-9
# define MOD 100000007
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=905;
//Code begin...

struct qnode{
    int v, c;
    qnode(int _v=0, int _c=0):v(_v),c(_c){}
    bool operator <(const qnode &r)const{return c>r.c;}
};
struct Edge{int p, next, w;}edge[N*8];
int head[N], cnt=1, dis[N], G[N][N];
char s[35][35];
bool vis[N];

void add_edge(int u, int v, int w){
    edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;
}
void Dijkstra(int n, int start){
    mem(vis,0);
    FO(i,0,n) dis[i]=INF;
    priority_queue<qnode>que;
    while (!que.empty()) que.pop();
    dis[start]=0; que.push(qnode(start,0)); qnode tmp;
    while (!que.empty()) {
        tmp=que.top(); que.pop();
        int u=tmp.v;
        if (vis[u]) continue;
        vis[u]=true;
        for (int i=head[u]; i; i=edge[i].next) {
            int v=edge[i].p, cost=edge[i].w;
            if (!vis[v]&&dis[v]>dis[u]+cost) dis[v]=dis[u]+cost, que.push(qnode(v,dis[v]));
        }
    }
}
int main ()
{
    int n, m, T;
    scanf("%d%d%d",&n,&m,&T);
    FO(i,0,n) scanf("%s",s[i]);
    FO(i,0,n) FO(j,0,m) {
        if (j!=m-1) add_edge(i*m+j,i*m+j+1,s[i][j+1]=='1'), add_edge(i*m+j+1,i*m+j,s[i][j]=='1');
        if (i!=n-1) add_edge(i*m+j,i*m+j+m,s[i+1][j]=='1'), add_edge(i*m+j+m,i*m+j,s[i][j]=='1');
    }
    double ans=0;
    FO(i,0,n) FO(j,0,m) {
        Dijkstra(n*m,i*m+j);
        FO(k1,0,n) FO(k2,0,m) {
            G[i*m+j][k1*m+k2]=dis[k1*m+k2]+(s[i][j]=='1');
            if (G[i*m+j][k1*m+k2]>T) continue;
            ans=max(ans,sqrt((i-k1)*(i-k1)+(j-k2)*(j-k2)));
        }
    }
    printf("%.6f
",ans);
    return 0;
}