Tri_integral Autumn Training 5 zoj 3294 Utawarerumono by Hyoga
2013-10-08 21:00
453 查看
题目是可以做的,感觉这个题目就是:脑筋急转弯+预处理+网络流。比赛的时候队友其实已经想的差不多了,就是一直觉得边太多没敢去尝试。实际上当满足所有条件之后边非常少不超过10000。跑网络流完全可以承受。
1.bitcount[i]>11是个很有力的剪枝。这么一搞符合条件的数i、j只有3000多个了。
2.对于题目中说的这个条件“For a square formed by n*n unit squares, you can fill each grid with white or black color. If two filling designs match or match after rotation, they are considered the same. Then the number of different filling designs should
be greater than or equal to 2500002 “(脑筋急转弯)
可知n>=50002一定可以,n<50000一定不可以。n=50001不可判定,但题目要求n为偶数,所以也不可以。又由于n=i&j,于是题目中满足条件的i,j一定都>=50002。
1.bitcount[i]>11是个很有力的剪枝。这么一搞符合条件的数i、j只有3000多个了。
2.对于题目中说的这个条件“For a square formed by n*n unit squares, you can fill each grid with white or black color. If two filling designs match or match after rotation, they are considered the same. Then the number of different filling designs should
be greater than or equal to 2500002 “(脑筋急转弯)
可知n>=50002一定可以,n<50000一定不可以。n=50001不可判定,但题目要求n为偶数,所以也不可以。又由于n=i&j,于是题目中满足条件的i,j一定都>=50002。
#include<iostream> #include<vector> #include<cstring> #include<cstdio> #include<queue> using namespace std; const int maxn=210000; const int inf=1<<28; int bitcount[maxn]; int a[maxn],map[maxn]; int u[maxn],v[maxn]; int solve(int x) { int num=0; while(x!=0) { if((x&1)==1)num++; x=x>>1; } return num; } //dinic struct Edge { int from,to,cap,flow; }; struct Dinic { int n,m,s,t; vector<Edge>edges; vector<int>g[maxn]; int d[maxn],cur[maxn]; bool vis[maxn]; void init(int n) { m=0;this->n=n; edges.clear(); for(int i=0;i<=n;i++) g[i].clear(); } void addedge(int from,int to,int cap) { edges.push_back((Edge){from,to,cap,0}); edges.push_back((Edge){to,from,0,0}); m=edges.size(); g[from].push_back(m-2); g[to].push_back(m-1); } bool bfs() { memset(vis,0,sizeof(vis)); queue<int>q; q.push(s); d[s]=0; vis[s]=1; while(!q.empty()) { int x=q.front();q.pop(); for(int i=0;i<g[x].size();i++) { Edge & e=edges[g[x][i]]; if(!vis[e.to]&& e.cap>e.flow) { vis[e.to]=1; d[e.to]=d[x]+1; q.push(e.to); } } } return vis[t]; } int dfs(int x,int a) { if(x==t || a==0)return a; int flow=0,f; for(int & i=cur[x];i<g[x].size();i++) { Edge & e=edges[g[x][i]]; if(d[x]+1==d[e.to] && (f=dfs(e.to,min(a,e.cap-e.flow)))>0) { e.flow+=f; edges[g[x][i]^1].flow-=f; flow+=f; a-=f; if(a==0)break; } } return flow; } int maxflow(int s,int t) { this->s=s;this->t=t; int flow=0; while(bfs()) { memset(cur,0,sizeof(cur)); flow+=dfs(s,inf); } return flow; } }ans; int main() { //freopen("data","r",stdin); //pre for(int i=0;i<=100000;i++) { bitcount[i]=solve(i); map[i]=-1; } int tote=0;int tot=0; for(int i=50002;i<=100000;i++) if(bitcount[i]>11) { tot++; a[tot]=i; map[i]=tot; } for(int i=1;i<=tot;i++) for(int j=i+1;j<=tot;j++) if(bitcount[(a[i]&a[j])]>11 && ((a[i]&a[j])>50002) && (a[i]&a[j])%2==0) { tote++; u[tote]=i;v[tote]=j; //u tot v } int s,t; while(scanf("%d%d",&s,&t)!=EOF) { if(s==t) { printf("150\n"); continue; } if(map[s]==-1 || map[t]==-1 || s>t) { printf("0\n"); continue; } // s=map[s]; t=map[t]; ans.init(tot+1); //addedge ans.addedge(tot+1,s,150); s=tot+1; for(int i=1;i<=tote;i++) ans.addedge(u[i],v[i],1); int out=ans.maxflow(s,t); printf("%d\n",out); } return 0; }
相关文章推荐
- Tri_integral Autumn Training 2
- Tri_integral Autumn Training 3 训练总结
- Tri_integral Winter Training 0 比赛总结
- Tri_integral Summer Training 9 总结
- Tri_integral Summer Training 6
- Tri_integral Summer Training 2 总结
- Tri_integral Summer Training 5 总结
- Tri_integral Summer Training 5 总结
- Tri_integral Summer Training 8 总结
- Tri_integral Summer Training 1 (Petrozavodsk Summer 2005. Moscow+Ural Contest) 总结
- Tri_integral Summer Training 3 总结
- Tri_integral Summer Training 8 总结
- Tri_integral Summer Training 4 总结(Hsinchu 2011)
- Tri_integral Summer Training 7 总结
- Tri_integral Summer Training 9 总结
- HDU3836 Equivalent Sets 2011 Multi-University Training Contest 1 - Host by HNU
- ZOJ 142 - The 14th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple - D
- Batch Normalization: Accelerating Deep Network Training by Reducing Internal Covariate Shift 论文翻译(转)
- 2014 Multi-University Training Contest 2--by 镇海中学 解题报告
- 2010 ACM-ICPC Multi-University Training Contest(12)——Host by WHU 部分题目解题报告