What is it in the 2019 South China University of Technology? Is there any empty? Can you come to alchem? (Thinking of thinking)

https://ac.nowcoder.com/acm/contest/625/B

Analysis:

All status is only 1<< 18, so we can preprocess f[u][j], and then build a state diagram of all the states that u can transfer;

There is a priority queue to search

There is a pitfall to be noted here, the array f[i][j], yes Can not open to long long, this will burst the memory; but f[i][j-1]* f[i][j-1] This operation will burst int, so we need to change

< p>f[i][j-1]* f[i][j-1] becomes long long 1ll*f[i][j-1]* f[i][j-1] Then go to mod

#include

using namespace std; #define ll long long

const int maxn = 1<<18; const int mod = 19260817; const int INF = < span style="color: #800080;">0x3f3f3f3f; bool  vabook[maxn],vis[maxn]; int f[maxn][20]; int  fa[maxn]; int dis[maxn]; struct no {int v; int val; bool operator <(const no &x) const {return< /span> x.val<val;} }P; vectorG[maxn]; priority_queue que; int main() {int n;scanf("%d",& n); bool T=< span style="color: #800080;">0; int  x; for(int< /span> i=1; i<=n; i++) {scanf(< span style="color: #800000;">"%d",&x); if(vabook[x]) T=1; vabook[x]=1< /span>;} if(T) {puts("0");return 0;} /// If the given number repeats, it must be 0 better



 for(int i=0; i/// preprocess all F[i][j]

 {f[i][0]=i; for(int j= 1; j<18; j++)/// a^4=a^2 * a^2, a^8=a^4*a^4;

 {f[i][j] = (1ll*f[i][j-1]* f[i][j-1]) %mod;/ // first convert to long long and then go to mod

}} for(int u=0; u) { for(int j=0 ; j<18; j++) {int v=u^(1<<j); x = max(u,v); G[u].push_back({v, f[x][j]+1});///Create one Picture, record the status and value of u can be transferred

} if(vabook[u])///This is the point I own

 {fa[u]=u; dis[u]=< span style="color: #800080;">0; que.push({u,0});} else {fa[u]=-1; dis[u ]=INF;}} ll ans=0x3f3f3f3f; //DJ shortest path

 while(!que.empty()) {P=que.top(); que.pop(); int u=Pv; if(P.val>=ans)  break;///If the smallest value of the current state is less than ans, then No need to update it

 if(vis[u]) continue; vis[u]=1;  for(int i=0 ; i) {int v=G[u][i].v; int w=< span style="color: #000000;">G[u][i].val; if(fa[v]!=- 1 && fa[v]!=fa[u]) {ans=min( ans,1ll*(dis[v]+w+dis[u])); //ans%=mod;

} if(dis[v]> w+dis[u]) {dis[v]=w+dis[u]; fa[v]=< span style="color: #000000;">fa[u]; que.push({v,dis[v]});}}} printf("%lld
",ans%mod); return 0< span style="color: #000000;">; }

View Code< /span>

#include

using namespace std; #define ll long long

const int maxn = 1<<18; const int mod = 19260817; const int INF = < span style="color: #800080;">0x3f3f3f3f; bool  vabook[maxn],vis[maxn]; int f[maxn][20]; int  fa[maxn]; int dis[maxn]; struct no {int v; int val; bool operator <(const no &x) const {return< /span> x.val<val;} }P; vectorG[maxn]; priority_queue que; int main() {int n;scanf("%d",& n); bool T=< span style="color: #800080;">0; int  x; for(int< /span> i=1; i<=n; i++) {scanf(< span style="color: #800000;">"%d",&x); if(vabook[x]) T=1; vabook[x]=1< /span>;} if(T) {puts("0");return 0;} /// If the given number repeats, it must be 0 better



 for(int i=0; i/// preprocess all F[i][j]

 {f[i][0]=i; for(int j= 1; j<18; j++)/// a^4=a^2 * a^2, a^8=a^4*a^4;

 {f[i][j] = (1ll*f[i][j-1]* f[i][j-1]) %mod;/ // first convert to long long and then go to mod

}} for(int u=0; u) { for(int j=0 ; j<18; j++) {int v=u^(1<<j); x = max(u,v); G[u].push_back({v, f[x][j]+1});///Create one Picture, record the status and value of u can be transferred

} if(vabook[u])///This is the point I own

 {fa[u]=u; dis[u]=< span style="color: #800080;">0; que.push({u,0});} else {fa[u]=-1; dis[u ]=INF;}} ll ans=0x3f3f3f3f; //DJ shortest path

 while(!que.empty()) {P=que.top(); que.pop(); int u=Pv; if(P.val>=ans)  break;///If the smallest value of the current state is less than ans, then No need to update it

 if(vis[u]) continue; vis[u]=1;  for(int i=0 ; i) {int v=G[u][i].v; int w=< span style="color: #000000;">G[u][i].val; if(fa[v]!=- 1 && fa[v]!=fa[u]) {ans=min( ans,1ll*(dis[v]+w+dis[u])); //ans%=mod;

} if(dis[v]> w+dis[u]) {dis[v]=w+dis[u]; fa[v]=< span style="color: #000000;">fa[u]; que.push({v,dis[v]});}}} printf("%lld
",ans%mod); return 0< span style="color: #000000;">; }

View Code< /span>

#include

using namespace std; #define ll long long

const int maxn = 1<<18; const int mod = 19260817; const int INF = < span style="color: #800080;">0x3f3f3f3f; bool  vabook[maxn],vis[maxn]; int f[maxn][20]; int  fa[maxn]; int dis[maxn]; struct no {int v; int val; bool operator <(const no &x) const {return< /span> x.val<val;} }P; vectorG[maxn]; priority_queue que; int main() {int n;scanf("%d",& n); bool T=< span style="color: #800080;">0; int  x; for(int< /span> i=1; i<=n; i++) {scanf(< span style="color: #800000;">"%d",&x); if(vabook[x]) T=1; vabook[x]=1< /span>;} if(T) {puts("0");return 0;} /// If the given number repeats, it must be 0 better



 for(int i=0; i/// preprocess all F[i][j]

 {f[i][0]=i; for(int j= 1; j<18; j++)/// a^4=a^2 * a^2, a^8=a^4*a^4;

 {f[i][j] = (1ll*f[i][j-1]* f[i][j-1]) %mod;/ // first convert to long long and then go to mod

}} for(int u=0; u) { for(int j=0 ; j<18; j++) {int v=u^(1<<j); x = max(u,v); G[u].push_back({v, f[x][j]+1});///Create one Picture, record the status and value of u can be transferred

} if(vabook[u])///This is the point I own

 {fa[u]=u; dis[u]=< span style="color: #800080;">0; que.push({u,0});} else {fa[u]=-1; dis[u ]=INF;}} ll ans=0x3f3f3f3f; //DJ shortest path

 while(!que.empty()) {P=que.top(); que.pop(); int u=Pv; if(P.val>=ans)  break;///If the smallest value of the current state is less than ans, then No need to update it

 if(vis[u]) continue; vis[u]=1;  for(int i=0 ; i) {int v=G[u][i].v; int w=< span style="color: #000000;">G[u][i].val; if(fa[v]!=- 1 && fa[v]!=fa[u]) {ans=min( ans,1ll*(dis[v]+w+dis[u])); //ans%=mod;

} if(dis[v]> w+dis[u]) {dis[v]=w+dis[u]; fa[v]=< span style="color: #000000;">fa[u]; que.push({v,dis[v]});}}} printf("%lld
",ans%mod); return 0< span style="color: #000000;">; }