XVIII Open Cup named after E.V. Pankratiev. GP of Urals

A. Nutella’s Life

斜率优化DP显然，CDQ分治后按$a$排序建线段树，每层维护凸包，查询时不断将队首弹出即可。

时间复杂度$O(n\log^2n)$。

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
const int N=100010,M=262150;
int n,i,a[N],cb;ll f[N],g[N],w[N],ans;
int st[M],en[M],tot,q[4000000];
int H;
P b[N];
inline void up(ll&a,ll b){a<b?(a=b):0;}
inline double pos(int x,int y){return 1.0*(g[x]-g[y])/(y-x);}
inline void push(int x){while(H<tot&&pos(q[tot-1],q[tot])>pos(q[tot],x))tot--;q[++tot]=x;
}
void build(int x,int l,int r){if(l==r){st[x]=en[x]=++tot;q[tot]=b[l].second;return;}int mid=(l+r)>>1;build(x<<1,l,mid);build(x<<1|1,mid+1,r);int i=st[x<<1],j=st[x<<1|1];H=st[x]=tot+1;while(i<=en[x<<1]&&j<=en[x<<1|1]){if(q[i]<q[j]){push(q[i++]);}else{push(q[j++]);}}while(i<=en[x<<1])push(q[i++]);while(j<=en[x<<1|1])push(q[j++]);en[x]=tot;
}
inline ll cal(int x,int y){return g[y]+2LL*x*y;}
void ask(int x,int a,int b,int d,int p){if(b<=d){while(st[x]<en[x]&&cal(p,q[st[x]])<cal(p,q[st[x]+1]))st[x]++;if(st[x]<=en[x])up(f[p],cal(p,q[st[x]])+w[p]);return;}int mid=(a+b)>>1;ask(x<<1,a,mid,d,p);if(d>mid)ask(x<<1|1,mid+1,b,d,p);
}
void solve(int l,int r){if(l==r){g[l]=f[l]-l-1LL*l*l;up(ans,f[l]-1LL*(n-l)*(n-l+1));return;}int mid=(l+r)>>1,i;solve(l,mid);cb=0;for(i=l;i<=mid;i++)b[++cb]=P(a[i],i);sort(b+1,b+cb+1);tot=0;build(1,1,cb);for(i=mid+1;i<=r;i++){if(a[i]<b[1].first)continue;int t=lower_bound(b+1,b+cb+1,P(a[i],i))-b-1;ask(1,1,cb,t,i);}solve(mid+1,r);
}
int main(){scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&a[i]),a[i]*=2;ans=-1LL*n*(n+1);for(i=1;i<=n;i++){f[i]=-1LL*(i-1)*i+a[i];w[i]=-1LL*i*i+i+a[i];}solve(1,n);printf("%lld",ans/2);
}

B. Oleg and Data Science

若$R<Q$，那么显然$\bmod Q$操作无效，故答案为无穷。

否则若$\lfloor\frac{L}{Q}\rfloor=\lfloor\frac{R}{Q}\rfloor$，则答案为$Q\lfloor\frac{L}{Q}\rfloor$的因子个数。

否则答案为$Q$的因子个数。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
LL L, R, Q;
LL check(LL n)
{LL rtn = 1;for(LL x = 2; x * x <= n; ++x)if(n % x == 0){int cnt = 0;while(n % x == 0){n /= x;++cnt;}rtn *= cnt + 1;}if(n >= 2)rtn *= 2;return rtn;
}
LL bf(bool out)
{int rtn = 0;for(int X = 1; X <= 40; ++X){  bool flag = 1;for(int S = L; S <= R; ++S){if(S % Q % X != S % X){flag = 0;break;}}if(flag){if(out)printf("%d ", X);}rtn += flag;}if(out)puts("END");return rtn;
}
int main()
{while(~scanf("%lld%lld%lld",&L,&R,&Q))//while(R = rand() % 30 + 1, L = rand() % R + 1, Q = rand() % 30 + 1, true){if(R<Q){puts("infinity");}else{//LL ans = check(max(Q, L / Q * Q));LL ans;if(L / Q == R / Q)ans = check(L / Q * Q);else ans = check(Q);printf("%lld\n", ans);/*if(ans != bf(0)){printf("%lld %lld %lld: %lld %lld\n", L, R, Q, ans, bf(1));getchar();//while(1);}*/}}return 0;
}/*
【trick&&吐槽】【题意】【分析】【时间复杂度&&优化】*/

C. Christmas Garland

首先将相邻同色块合并，建立一张图，每个点表示一种颜色的灯，两点间的边表示这两种颜色相邻的有多少个，那么答案为亮着的灯数减去同时亮着的边数。

将点按照度数大小以$\sqrt{m}$为界分成小点和大点，对于每个点维护$w_x$表示周围有多少个小点亮着。

那么若操作小点，则枚举它所有相连的边，修改其$w$，同时对于相邻的大点计算对答案的贡献。

若操作大点，则枚举它所有与大点相连的边，计算贡献。

时间复杂度$O(n\sqrt{n})$。

#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;
typedef pair<int,int>P;
const int N=200010;
int n,ce,_,m,Q,i,j,x,y,have[N],tot,ans,lim;
int d[N],a[N],f[N],w[N];
vector<P>g[N],gg[N];
vector<P>::iterator it;
struct E{int x,y;}e[N];
inline bool cmp(const E&a,const E&b){return a.x==b.x?a.y<b.y:a.x<b.x;}
inline void add(int x,int y,int z){d[x]++;lim++;g[x].push_back(P(y,z));
}
inline void addd(int x,int y,int z){if(d[x]>=lim&&d[y]>=lim)gg[x].push_back(P(y,z));
}
int main(){scanf("%d%d%d",&_,&m,&Q);while(_--){scanf("%d",&x);if(x!=a[n])a[++n]=x;}for(i=1;i<=n;i++)have[a[i]]++;for(i=1;i<n;i++){x=a[i],y=a[i+1];if(x>y)swap(x,y);e[++ce].x=x;e[ce].y=y;}sort(e+1,e+ce+1,cmp);for(i=1;i<=ce;i=j){for(j=i;j<=ce&&e[i].x==e[j].x&&e[i].y==e[j].y;j++);add(e[i].x,e[i].y,j-i);add(e[i].y,e[i].x,j-i);}lim=sqrt(lim);for(i=1;i<=ce;i=j){for(j=i;j<=ce&&e[i].x==e[j].x&&e[i].y==e[j].y;j++);addd(e[i].x,e[i].y,j-i);addd(e[i].y,e[i].x,j-i);}while(Q--){scanf("%d",&x);if(d[x]<lim){if(f[x]==0){tot+=have[x];ans+=w[x];for(it=g[x].begin();it!=g[x].end();it++){w[it->first]+=it->second;if(d[it->first]>=lim&&f[it->first])ans+=it->second;}}else{tot-=have[x];ans-=w[x];for(it=g[x].begin();it!=g[x].end();it++){w[it->first]-=it->second;if(d[it->first]>=lim&&f[it->first])ans-=it->second;}}}else{if(f[x]==0){tot+=have[x];ans+=w[x];for(it=gg[x].begin();it!=gg[x].end();it++)if(f[it->first])ans+=it->second;}else{tot-=have[x];ans-=w[x];for(it=gg[x].begin();it!=gg[x].end();it++)if(f[it->first])ans-=it->second;}}f[x]^=1;printf("%d\n",tot-ans);}
}

D. Anna and Lucky Tickets

问题即统计有多少长度为$n$的回文十进制数字串满足奇数位数位和等于偶数位数位和。

若$n$是偶数，那么因为回文的性质，左半部分在右半部分奇偶性反转，故奇数位和必定等于偶数位和，故$ans=10^{\frac{n}{2}}$。

若$n$是奇数，那么奇数位要填$[0,9]$，偶数位要填$[-9,0]$，将所有偶数位都加上$9$，再枚举中间的数，则问题转化为统计$\sum_{i=1}^n x_i=m,0\leq x_i\leq 9$的解数。

考虑容斥，枚举有$k$个$x_i$必定突破了上限，剩下的不管，将上限从$m$中减掉后用隔板法计算，则$ans=\sum_{k=0}^n (-1)^kC(n,k)C(m-10k+n-1,n-1)$。

时间复杂度$O(n)$。

#include<cstdio>
const int N=10000010,P=1000000007;
int n,i,ans,even,odd,all,fac[N],inv[N];
inline int po(int a,int b){int t=1;for(;b;b>>=1,a=1LL*a*a%P)if(b&1)t=1LL*t*a%P;return t;}
inline int C(int n,int m){if(n<m)return 0;if(n==m)return 1;return 1LL*fac[n]*inv[m]%P*inv[n-m]%P;
}
int cal(int sum,int n){int ret=0;for(int i=0;i<=n;i++){int t=1LL*C(sum-i*10+n-1,n-1)*C(n,i)%P;if(i&1)ret=(ret-t+P)%P;else ret=(ret+t)%P;}return ret;
}
int main(){scanf("%d",&n);if(n%2==0)return printf("%d",po(10,n/2)),0;for(fac[0]=i=1;i<N;i++)fac[i]=1LL*fac[i-1]*i%P;for(inv[0]=inv[1]=1,i=2;i<N;i++)inv[i]=1LL*(P-inv[P%i])*(P/i)%P;for(i=2;i<N;i++)inv[i]=1LL*inv[i-1]*inv[i]%P;even=n/2;odd=n-even;all=even*9;if((n+1)/2%2)odd--;else even--;odd/=2;even/=2;for(i=0;i<10;i++){int t=all-i;if(t<0||t%2)continue;ans=(cal(t/2,odd+even)+ans)%P;}printf("%d",ans);
}

E. Octopus

分类大讨论。

F. Secret Permutation

首先进行一轮迭代$Q(zero,i,zero)$即可找到$0$的位置。

将剩下$n-1$个数放入集合$S$，每次从$S$中随机选择两个$x,y$，询问$Q(x,y,zero)$，即可有很大概率排除掉$x$和$y$，期望$\frac{n}{2}$次可以得到$1$的位置。

那么知道$1$之后，不断询问$Q(i,one,one)$即可得到所有数的位置。

询问次数期望为$2.5n\leq 12512$。

#include<cstdio>
#include<cstdlib>
#include<ctime>
#include<algorithm>
using namespace std;
const int N=5010;
int n,i,A,B,x,m,q[N],f[N],one,zero;
inline int ask(int a,int b,int c){printf("? %d %d %d\n",a,b,c);fflush(stdout);int t;scanf("%d",&t);return t;
}
int main(){srand(time(NULL));scanf("%d",&n);if(n==1){puts("! 0");fflush(stdout);return 0;}for(i=0;i<n;i++)zero=ask(zero,i,zero);for(i=0;i<n;i++)if(i!=zero)q[m++]=i;while(m>1){A=q[x=rand()%m];swap(q[x],q[--m]);B=q[x=rand()%m];swap(q[x],q[--m]);x=ask(A,B,zero);if(x==A)q[m++]=B;if(x==B)q[m++]=A;}f[A=one=q[0]]=1;for(i=2;i<n;i++)f[A=ask(A,one,one)]=i;printf("!");for(i=0;i<n;i++)printf(" %d",f[i]);fflush(stdout);
}

G. Polygon Rotation

留坑。

H. Mikhail’s Problem

论文题。回文树求出每个回文等差数列消失的位置，树状数组维护，时间复杂度$O(n\log n)$。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int N=100010;
int sz,last,go[N][26],len[N],diff[N],link[N],series_link[N],baby[N];
int n,m,i,j,k,x,y,f[N],bit[N];
int ans[N];
vector<pair<int,int> >poses[N],que[N];
char s[N];
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
inline int go_until_pal(int i,int v){while(s[i]!=s[i-len[v]-1])v=link[v];return v;
}
inline pair<vector<int>,vector<int> >add_char(int i){last=go_until_pal(i,last);int&v=go[last][s[i]-'a'];if(v==-1){v=sz++;len[v]=len[last]+2;if(!last)link[v]=1;else{last=go_until_pal(i,link[last]);link[v]=go[last][s[i]-'a'];}diff[v]=len[v]-len[link[v]];if(diff[v]==diff[link[v]]){baby[v]=baby[link[v]];series_link[v]=series_link[link[v]];}else{baby[v]=v;series_link[v]=link[v];}}last=v;pair<vector<int>,vector<int> >ans;for(int u=v;u!=1;u=series_link[u]){int B=baby[u];vector<pair<int,int> >&vec=poses[B];int left=i-len[u]+1;if(vec.size()&&vec.back().first==left)vec.back().second=len[u];else vec.push_back(make_pair(left,len[u]));ans.first.push_back(i-len[B]+1);if(vec.size()!=1){int L=vec[vec.size()-2].first,len_=vec[vec.size()-2].second,R=L+len_-1,L_=R-len[u]+1;ans.second.push_back(L_);}if(vec.size()!=1)if(vec[vec.size()-1].second==vec[vec.size()-2].second){vec[vec.size()-2]=vec[vec.size()-1];vec.pop_back();}}return ans;
}
inline void add(int x,int p){for(f[x]+=p;x<=n;x+=x&-x)bit[x]+=p;}
inline int ask(int x){int t=0;for(;x;x-=x&-x)t+=bit[x];return t;}
int main(){scanf("%s%d",s+1,&m);n=strlen(s+1);sz=2,last=0,len[0]=-1;for(i=0;i<=n+5;i++)memset(go[i],-1,sizeof(go[i]));for(i=0;i<m;i++){read(x),read(y);que[y].push_back(make_pair(x,i));}for(i=1;i<=n;i++){pair<vector<int>,vector<int> >vec=add_char(i);vector<int>adding=vec.first,deleting=vec.second; for(j=0;j<deleting.size();j++){if(!f[k=deleting[j]])continue;add(k,-1);}for(j=0;j<adding.size();j++){if(f[k=adding[j]]==1)continue;add(k,1);}for(j=0;j<que[i].size();j++)ans[que[i][j].second]=sz-2-ask(que[i][j].first-1);}for(i=0;i<m;i++)printf("%d\n",ans[i]);
}

I. Rat-O-Matic

若$A$包含$B$，则$B$的半径不超过$A$的一半，故扫描线将包含关系建树后，树高不超过$O(\log R)$。

对于每个字符串，枚举所有后缀，将长度$O(\log R)$的前缀插入字典树，即可完成询问。

J. Readability

贪心，求出每个数的活动区间，扫描线的时候将最优的那个数填下来即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
int a[N], b[N], c[N];
int v[2];
int w[2][N];
int P(int o, int id)
{return id * 2 - o;
}
struct A
{int pos, val;
};
bool gorgt(A a, A b)
{if(a.pos != b.pos)return a.pos < b.pos;return a.val > b.val;
}
bool golft(A a, A b)
{if(a.pos != b.pos)return a.pos > b.pos;return a.val > b.val;
}
void GORGT(int st, int ed, int val, int o, int aimo, int b[])
{vector<A>vt;for(int i = st; ; i += val){vt.push_back({w[o][i], 1});vt.push_back({P(aimo,i), -1});if(i == ed)break;}sort(vt.begin(), vt.end(), gorgt);multiset<int>sot;for(auto it : vt){if(it.val == 1){sot.insert(a[it.pos]);}else{b[it.pos] = *sot.begin();sot.erase(sot.begin());}}
}
void GOLFT(int st, int ed, int val, int o, int aimo, int b[])
{vector<A>vt;for(int i = st; ; i += val){vt.push_back({w[o][i], 1});vt.push_back({P(aimo,i), -1});if(i == ed)break;}sort(vt.begin(), vt.end(), golft);multiset<int>sot;for(auto it : vt){if(it.val == 1){sot.insert(a[it.pos]);}else{b[it.pos] = *--sot.end();sot.erase(--sot.end());}}
}
LL solve(int o, int aimo, int b[])
{LL rtn = 0;//try to check every numfor(int i = 1; i <= v[o]; ++i){int l = w[o][i];    //start posint r = P(aimo, i); //end posrtn += abs(l - r);if(l == r){b[r] = a[l];}else if(l < r){int p = i;while(p < v[o]){int ll = w[o][p + 1];int rr = P(aimo, p + 1);if(ll <= r){r = rr;++p;rtn += abs(ll - rr);}else break;}GORGT(i, p, 1, o, aimo, b);i = p;}else{swap(l, r);int p = i;while(p < v[o]){int ll = w[o][p + 1];int rr = P(aimo, p + 1);swap(ll, rr);if(ll <= r){r = rr;++p;rtn += abs(ll - rr);}else break;}GOLFT(i, p, 1, o, aimo, b);i = p;}}return rtn;
}
void print(int b[])
{for(int i = 1; i <= n; ++i){printf("%d%c", b[i], i == n ? '\n' : ' ');}
}
bool small(int b[], int c[])
{for(int i = 1; i <= n; ++i){if(b[i] > c[i])return 0;if(b[i] < c[i])return 1;}return 0;
}
int p[N];
int d[N], ansd[N];
LL bf()
{int ans = 1e9;for(int i = 1; i <= n; ++i)p[i] = i;do{int tmp = 0;bool flag = 1;for(int i = 1; i <= n; ++i){tmp += abs(p[i] - i);d[i] = a[p[i]];if(i >= 2 && d[i] % 2 == d[i - 1] % 2)flag = 0;}if(flag){if(tmp < ans || tmp == ans && small(d, ansd)){ans = tmp;for(int i = 1; i <= n; ++i)ansd[i] = d[i];}}}while(next_permutation(p + 1, p + n + 1));
}
int main()
{while(~scanf("%d",&n))//while(n = rand() % 9 + 1, true){//for(int i = 1; i <= n; ++i)a[i] = i; random_shuffle(a + 1, a + n + 1);//printf("A: ");print(a);v[0] = v[1] = 0;for(int i = 1; i <= n; ++i){scanf("%d", &a[i]);if(a[i] & 1){w[1][++v[1]] = i;}else{w[0][++v[0]] = i;}}//bf();if(n & 1){//puts("n & 1");//偶 奇 偶if(v[0] > v[1]){solve(0, 1, b);solve(1, 0, b);}//奇 偶 奇else{solve(1, 1, b);solve(0, 0, b);}print(b);/*if(small(ansd, b)){puts("NO");print(ansd);getchar();}*/}else{//ans1 偶 奇LL ans1 = solve(0, 1, b) + solve(1, 0, b);//ans2 ji ouLL ans2 = solve(1, 1, c) + solve(0, 0, c);if(ans1 < ans2){print(b);/*if(small(ansd, b)){puts("NO");print(ansd);getchar();}*/}else if(ans1 > ans2){print(c);/*if(small(ansd, c)){puts("NO");print(ansd);getchar();}*/}else{if(small(b, c)){print(b);/*if(small(ansd, b)){puts("NO");print(ansd);getchar();}*/}else{print(c);/*if(small(ansd, c)){puts("NO");print(ansd);getchar();}*/}}}}return 0;
}/*
【trick&&吐槽】
4
1 3 2 44
3 1 4 2【题意】【分析】【时间复杂度&&优化】*/

K. Robotobor

BFS出$d[i][j][x][y]$表示从$(i,j)$到$(x,y)$的最短回文路线。

第二次BFS出$f[i][j]$表示从$(i,j)$到终点最少需要多少条长度不超过$100$的回文路线。

时间复杂度$O(n^4)$。

#include<cstdio>
const int N=55;
int n,m,i,j,x,y;
char a[N][N];
int d[N][N][N][N],f[N][N];
char q[N*N*N*N][4];
int h,t;
inline void ext(int A,int B,int C,int D,int w){if(A<1||A>n||B<1||B>m)return;if(a[A][B]=='#')return;if(C<1||C>n||D<1||D>m)return;if(a[C][D]=='#')return;if(~d[A][B][C][D])return;q[++t][0]=A;q[t][1]=B;q[t][2]=C;q[t][3]=D;d[A][B][C][D]=w;
}
inline void up(int x,int y,int w){if(~f[x][y])return;q[++t][0]=x;q[t][1]=y;f[x][y]=w;
}
inline int abs(int x){return x>0?x:-x;}
void print(int A,int B,int C,int D){if(A==C&&B==D)return;if(abs(A-C)+abs(B-D)==1){if(A+1==C)putchar('D');if(A-1==C)putchar('U');if(B+1==D)putchar('R');if(B-1==D)putchar('L');return;}if(~d[A][B-1][C][D+1]&&d[A][B-1][C][D+1]+2==d[A][B][C][D]){putchar('L');print(A,B-1,C,D+1);putchar('L');return;}if(~d[A][B+1][C][D-1]&&d[A][B+1][C][D-1]+2==d[A][B][C][D]){putchar('R');print(A,B+1,C,D-1);putchar('R');return;}if(~d[A-1][B][C+1][D]&&d[A-1][B][C+1][D]+2==d[A][B][C][D]){putchar('U');print(A-1,B,C+1,D);putchar('U');return;}if(~d[A+1][B][C-1][D]&&d[A+1][B][C-1][D]+2==d[A][B][C][D]){putchar('D');print(A+1,B,C-1,D);putchar('D');return;}puts("Error");while(1);
}
void show(int x,int y){while(f[x][y]){bool flag=0;for(i=1;i<=n;i++)for(j=1;j<=m;j++){if(flag)break;if(~d[x][y][i][j]&&d[x][y][i][j]<=100&&f[i][j]+1==f[x][y]){print(x,y,i,j);puts("");x=i,y=j;flag=1;break;}}}
}
int main(){scanf("%d%d",&n,&m);for(i=1;i<=n;i++)scanf("%s",a[i]+1);for(i=0;i<=n+1;i++)for(j=0;j<=m+1;j++)for(x=0;x<=n+1;x++)for(y=0;y<=m+1;y++){d[i][j][x][y]=-1;}h=1,t=0;for(i=1;i<=n;i++)for(j=1;j<=m;j++)ext(i,j,i,j,0);for(i=1;i<=n;i++)for(j=1;j<=m;j++){ext(i,j,i+1,j,1);ext(i,j,i-1,j,1);ext(i,j,i,j-1,1);ext(i,j,i,j+1,1);}while(h<=t){int A=q[h][0],B=q[h][1],C=q[h][2],D=q[h][3];h++;int w=d[A][B][C][D]+2;//Dext(A-1,B,C+1,D,w);//Uext(A+1,B,C-1,D,w);//Lext(A,B+1,C,D-1,w);//Rext(A,B-1,C,D+1,w);}h=1,t=0;for(i=1;i<=n;i++)for(j=1;j<=m;j++)f[i][j]=-1;for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(a[i][j]=='F')up(i,j,0);while(h<=t){int A=q[h][0],B=q[h][1];h++;int w=f[A][B]+1;for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(~d[i][j][A][B]&&d[i][j][A][B]<=100)up(i,j,w);}for(i=1;i<=n;i++)for(j=1;j<=m;j++)if(a[i][j]=='S'){if(f[i][j]<0)return puts("-1"),0;printf("%d\n",f[i][j]);show(i,j);return 0;}
}

转载于:https://www.cnblogs.com/clrs97/p/8667455.html

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