#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define maxn 50005 #define MAXN 100005 #define mod 1000000009 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 typedef long long ll; using namespace std; int n,m,ans,cnt,tot,flag; int len,st,ed; // 直徑 起點 終點 bool vis[maxn]; int head[maxn],val[maxn],dist[maxn][2]; int f[maxn][20],g[maxn][20],lg[maxn]; //第二維為二進位 struct Node { int v,w,next; } edge[MAXN]; void addedge(int u,int v,int w) { cnt++; edge[cnt].v=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt; } void dfs(int u,int fa,int sum){ if(sum>len){ len=sum; st=u; } int i,j,v; for(i=head[u];i;i=edge[i].next){ v=edge[i].v; if(v!=fa){ dfs(v,u,sum+edge[i].w); } } } void dfs1(int u,int fa,int k,int sum){ dist[u][k]=sum; int i,j,v; for(i=head[u];i;i=edge[i].next){ v=edge[i].v; if(v!=fa) dfs1(v,u,k,sum+edge[i].w); } } void init_rmq() // 預處理 O(n*log(n)) { int i,j; for(i=1;i<=n;i++) { f[i][0]=g[i][0]=val[i]; } for(j=1;(1<<j)<=n;j++) { for(i=1;i+j-1<=n;i++) { if((i+(1<<(j-1))<=n)) { f[i][j]=max(f[i][j-1],f[i+(1<<(j-1))][j-1]); //避免越界 g[i][j]=min(g[i][j-1],g[i+(1<<(j-1))][j-1]); } else { f[i][j]=f[i][j-1]; g[i][j]=g[i][j-1]; } } } } int query_rmq(int l,int r) { int k=lg[r-l+1]; return max(f[l][k],f[r-(1<<k)+1][k])-min(g[l][k],g[r-(1<<k)+1][k]); } int main() { int i,j,t; lg[1]=0; for(i=2;i<=50000;i++) { lg[i]=lg[i>>1]+1; } while(~scanf("%d%d",&n,&m)) { if(n==0&&m==0) break ; cnt=0; memset(head,0,sizeof(head)); int u,v,w; for(i=1;i<n;i++) { scanf("%d%d%d",&u,&v,&w); addedge(u,v,w); addedge(v,u,w); } len=0; dfs(1,0,0); ed=st; len=0; dfs(st,0,0); dfs1(st,0,0,0); dfs1(ed,0,1,0); for(i=1;i<=n;i++) { val[i]=max(dist[i][0],dist[i][1]); } init_rmq(); int q,le,ri,mid,res,id; while(m--){ scanf("%d",&q); ans=id=1; for(i=1;i<=n;i++) { while(id<=i){ res=query_rmq(id,i); if(res>q) id++; else break ; } ans=max(ans,i-id+1); } printf("%d ",ans); } } return 0; }
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> using namespace std; int val[255][255]; int mm[255]; int dpmin[255][255][8][8];//最小值 int dpmax[255][255][8][8];//最大值 void initRMQ(int n,int m) { for(int i = 1;i <= n;i++) for(int j = 1;j <= m;j++) dpmin[i][j][0][0] = dpmax[i][j][0][0] = val[i][j]; for(int ii = 0; ii <= mm[n]; ii++) for(int jj = 0; jj <= mm[m]; jj++) if(ii+jj) for(int i = 1; i + (1<<ii) - 1 <= n; i++) for(int j = 1; j + (1<<jj) - 1 <= m; j++) { if(ii) { dpmin[i][j][ii][jj] = min(dpmin[i][j][ii-1][jj],dpmin[i+(1<<(ii-1))][j][ii-1][jj]); dpmax[i][j][ii][jj] = max(dpmax[i][j][ii-1][jj],dpmax[i+(1<<(ii-1))][j][ii-1][jj]); } else { dpmin[i][j][ii][jj] = min(dpmin[i][j][ii][jj-1],dpmin[i][j+(1<<(jj-1))][ii][jj-1]); dpmax[i][j][ii][jj] = max(dpmax[i][j][ii][jj-1],dpmax[i][j+(1<<(jj-1))][ii][jj-1]); } } } //查询矩形的最大值 int rmq1(int x1,int y1,int x2,int y2) { int k1 = mm[x2-x1+1]; int k2 = mm[y2-y1+1]; x2 = x2 - (1<<k1) + 1; y2 = y2 - (1<<k2) + 1; return max(max(dpmax[x1][y1][k1][k2],dpmax[x1][y2][k1][k2]),max(dpmax[x2][y1][k1][k2],dpmax[x2][y2][k1][k2])); } //查询矩形的最小值 int rmq2(int x1,int y1,int x2,int y2) { int k1 = mm[x2-x1+1]; int k2 = mm[y2-y1+1]; x2 = x2 - (1<<k1) + 1; y2 = y2 - (1<<k2) + 1; return min(min(dpmin[x1][y1][k1][k2],dpmin[x1][y2][k1][k2]),min(dpmin[x2][y1][k1][k2],dpmin[x2][y2][k1][k2])); } int main() { mm[0] = -1; for(int i = 1;i <= 500;i++) mm[i] = ((i&(i-1))==0)?mm[i-1]+1:mm[i-1]; int N,B,K; for(int i=1;i<=250;i++){ if(i%5==0) printf(" "); printf("%10d",mm[i]); } while(scanf("%d%d%d",&N,&B,&K)==3) { for(int i = 1;i <= N;i++) for(int j = 1;j <= N;j++) scanf("%d",&val[i][j]); initRMQ(N,N); int x,y; while(K--) { scanf("%d%d",&x,&y); printf("%d ",rmq1(x,y,x+B-1,y+B-1)-rmq2(x,y,x+B-1,y+B-1)); } } return 0; }
#include <stdio.h> #include <string.h> #include <algorithm> #include <iostream> using namespace std; const int MAXN = 100010; int dp[MAXN][20]; int mm[MAXN]; void initRMQ(int n,int b[]) { mm[0] = -1; for(int i = 1;i <= n;i++) { mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1]; dp[i][0] = b[i]; } for(int j = 1;j <= mm[n];j++) for(int i = 1;i + (1<<j) - 1 <= n;i++) dp[i][j] = max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); } int rmq(int x,int y) { int k = mm[y-x+1]; return max(dp[x][k],dp[y-(1<<k)+1][k]); } int a[MAXN]; int hash[MAXN]; int L[MAXN],R[MAXN]; int b[MAXN]; int main() { int n,m; while(scanf("%d",&n) == 1 && n) { scanf("%d",&m); for(int i = 1;i <= n;i++) scanf("%d",&a[i]); int k = 1; int l = 1; memset(b,0,sizeof(b)); for(int i = 1;i <= n;i++) { if(i > 1 && a[i] != a[i-1]) { for(int j = l;j < i;j++) { L[j] = l; R[j] = i-1; } b[k] = i - l; l = i; k ++; } hash[i] = k; } for(int j = l;j <= n;j++) { L[j] = l; R[j] = n; } b[k] = n-l+1; initRMQ(k,b); int u,v; while(m--) { scanf("%d%d",&u,&v); int t1 = hash[u]; int t2 = hash[v]; if(t1 == t2) { printf("%d ",v-u+1); continue; } int ans = max(R[u]-u+1,v-L[v]+1); t1++; t2--; if(t1 <= t2)ans = max(ans,rmq(t1,t2)); printf("%d ",ans); } } return 0; }