标签：min 队列 cst 合并 namespace string break oid blog
A.
瞎jb折半dp一下，然后合并一下就好了。
O2加成十分显著。。。6s变成0.9s。。。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<bitset> #define maxv 100 #define maxe 20000 using namespace std; int n,m,d,x,y,z,g[maxv],nume=1,s1,s2,ans=0; struct edge { int v,f,nxt; }e[maxe]; bitset <maxv> dp1[maxv][(1<<10)+1][2],dp2[maxv][(1<<10)+1][2]; bitset <1030> vvv[maxv],kr; bool vis[(1<<20)+1],arr[maxv][15]; void addedge(int u,int v,int f) { e[++nume].v=v;e[nume].f=f; e[nume].nxt=g[u];g[u]=nume; } void dp1s() { for (register int i=1;i<=s1;i++) { for (register int j=1;j<=n;j++) for (register int k=0;k<(1<<s1);k++) dp1[j][k][i&1].reset(); for (register int j=1;j<=n;j++) for (register int k=g[j];k;k=e[k].nxt) { int v=e[k].v,f=e[k].f,rr=i&1; for (register int s=0;s<(1<<(i-1));s++) dp1[j][(s<<1)|f][rr]|=dp1[v][s][rr^1]; arr[j][i]|=arr[v][i-1]; } } } void dp2s() { for (register int i=1;i<=s2;i++) { for (register int j=1;j<=n;j++) for (register int k=0;k<(1<<s2);k++) dp2[j][k][i&1].reset(); for (register int j=1;j<=n;j++) for (register int k=g[j];k;k=e[k].nxt) { int v=e[k].v,f=e[k].f,rr=i&1; for (register int s=0;s<(1<<(i-1));s++) dp2[j][(s<<1)|f][rr]|=dp2[v][s][rr^1]; } } } void comb() { for (register int i=1;i<=n;i++) for (register int j=0;j<(1<<s1);j++) if (dp1[i][j][s1&1].count()) vvv[i][j]=1; for (register int i=1;i<=n;i++) for (register int j=0;j<(1<<s2);j++) { if (dp2[i][j][s2&1].count()) { kr.reset(); for (register int k=1;k<=n;k++) if (dp2[i][j][s2&1][k]) kr|=vvv[k]; for (register int k=0;k<(1<<s1);k++) { if (kr[k]) vis[k<<s2|j]=true; } } } } int main() { scanf("%d%d%d",&n,&m,&d); for (register int i=1;i<=m;i++) { scanf("%d%d%d",&x,&y,&z); addedge(x,y,z);addedge(y,x,z); } s1=d/2;s2=d-s1;arr[1][0]=true;dp1[1][0][0][1]=1; dp1s(); for (register int i=1;i<=n;i++) if (arr[i][s1]) dp2[i][0][0][i]=1; dp2s(); comb(); for (register int i=0;i<(1<<d);i++) ans+=vis[i]; printf("%d\n",ans); return 0; }
B.
考虑倒着做。这个答案只会单增，于是枚举答案然后check。
怎么check呢？考虑枚举所有这么长的包含这辆车的这一行中的区间，每个位置记下它上面和下面的第一个点（车）的位置，然后相当于区间max,min。
那么单调队列解决。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 2050 #define inf 2000000 using namespace std; int n,m,k,no[maxn][maxn],so[maxn][maxn],map[maxn][maxn],x[maxn],y[maxn],ans[maxn],dp[maxn][maxn]; int q1[maxn],h1,t1,q2[maxn],h2,t2; char s[maxn]; void get_table() { for (int j=1;j<=m;j++) { int ret=0; for (int i=1;i<=n;i++) { no[i][j]=ret; if (map[i][j]) ret=i; } ret=n+1; for (int i=n;i>=1;i--) { so[i][j]=ret; if (map[i][j]) ret=i; } } } void dp_() { for (int i=1;i<=n;i++) for (int j=1;j<=m;j++) { if (!map[i][j]) dp[i][j]=min(dp[i-1][j-1],min(dp[i-1][j],dp[i][j-1]))+1; else dp[i][j]=0; ans[k]=max(ans[k],dp[i][j]); } } bool check(int l,int x,int y) { int lb=-1,rb=inf,ret=0; for (int i=y-1;i>=0;i--) if (map[x][i]) {lb=i+1;break;} for (int i=y+1;i<=m+1;i++) if (map[x][i]) {rb=i-1;break;} int left=max(lb,y-l+1),right=min(y,rb-l+1);h1=h2=1;t1=t2=0; for (int i=left;i<=left+l-1;i++) { while (h1<=t1 && no[x][i]>=no[x][q1[t1]]) t1--;q1[++t1]=i; while (h2<=t2 && so[x][i]<=so[x][q2[t2]]) t2--;q2[++t2]=i; } for (int i=left;i<=right;i++) { ret=max(ret,so[x][q2[h2]]-no[x][q1[h1]]-1); while (h1<=t1 && q1[h1]<=i) h1++; while (h1<=t1 && no[x][i+l]>=no[x][q1[t1]]) t1--;q1[++t1]=i+l; while (h2<=t2 && q2[h2]<=i) h2++; while (h2<=t2 && so[x][i+l]<=so[x][q2[t2]]) t2--;q2[++t2]=i+l; } return ret>=l; } void ask() { for (int i=k;i>=2;i--) { map[x[i]][y[i]]=0; int ret=0; for (int j=1;j<=n;j++) { no[j][y[i]]=ret; if (map[j][y[i]]) ret=j; } ret=n+1; for (int j=n;j>=1;j--) { so[j][y[i]]=ret; if (map[j][y[i]]) ret=j; } int p=ans[i]+1; while (check(p,x[i],y[i]) && p<=min(n,m)) p++; ans[i-1]=p-1; } } int main() { scanf("%d%d%d",&n,&m,&k); for (int i=1;i<=n;i++) { scanf("%s",s+1); for (int j=1;j<=m;j++) { if (s[j]==‘X‘) map[i][j]=1; else map[i][j]=0; } } for (int i=1;i<=n;i++) map[i][0]=map[i][m+1]=1; for (int i=1;i<=m;i++) map[0][i]=map[n+1][i]=1; for (int i=1;i<=k;i++) { scanf("%d%d",&x[i],&y[i]); map[x[i]][y[i]]=1; } get_table(); dp_(); ask(); for (int i=1;i<=k;i++) printf("%d\n",ans[i]); return 0; }
C.
和上一场的C有什么区别吗。。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #define maxn 100500 using namespace std; long long n,prime[maxn],tot=0,w[maxn][60],ans=0,mn[maxn],f,kr=0; bool vis[maxn]; void get_table() { for (long long i=2;i<=maxn-500;i++) { if (!vis[i]) prime[++tot]=mn[i]=i; for (long long j=1;i*prime[j]<=maxn-500;j++) { vis[i*prime[j]]=true;mn[i*prime[j]]=prime[j]; if (i%prime[j]) continue;break; } } for (long long i=2;i<=maxn-500;i++) { long long ret=-1,x=i; while (x!=1) { if (mn[x]!=ret) {ret=mn[x];w[i][0]++;w[i][w[i][0]]=mn[x];} x/=mn[x]; } } } void dfs(long long now,long long top,long long ret,long long val,long long n) { if (now==w[top][0]+1) { if (!ret) return; if (ret&1) kr+=n/val;else kr-=n/val; return; } dfs(now+1,top,ret,val,n); dfs(now+1,top,ret+1,val*w[top][now],n); } void ask(long long n,long long val) {kr=0;dfs(1,val,0,1,n);} int main() { scanf("%lld",&n); get_table(); long long top=sqrt(n); for (long long i=2;i<=top;i++) { ask(n/i,i);ans+=(n/i)-kr; ask(i,i);ans-=i-kr; } printf("%lld\n",ans); return 0; }
标签：min 队列 cst 合并 namespace string break oid blog
原文地址：http://www.cnblogs.com/ziliuziliu/p/6411333.html
