题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1806
题目大意：给你一个非降序排列的整数数组，你的任务是对于一系列的询问，（i，j），回答序列中出现次数最多的数的个数；
如下图所示：
AC代码：
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<queue> 6 #include<string> 7 #include<cmath> 8 using namespace std; 9 const int N = 1e5+10; 10 int a[N],b[N]; 11 int dp[N][20]; 12 int n,m; 13 //构造和寻找的代码，来自大白书 14 void buildrmq( ) 15 { 16 for(int i=0;i<n;i++) 17 dp[i][0]=b[i]; 18 for(int j=1;(1<<j)<=n;j++) 19 for(int i=0;i+(1<<j)-1<n;i++) 20 dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); 21 } 22 int search(int s,int v) 23 { 24 int k =0 ; 25 while(1<<(k+1) <= v-s+1) k++; 26 return max(dp[s][k],dp[v-(1<<k)+1][k]); 27 } 28 int bi_search(int s,int t) 29 { 30 int tmp=a[t]; 31 int l=s; 32 int r=t; 33 int mid; 34 while(l<r) 35 { 36 mid=((l+r)>>1); 37 if(a[mid]>=tmp) r=mid; 38 else l=mid+1; 39 } 40 return r; 41 } 42 int main() 43 { 44 int T; 45 while(scanf("%d",&n) && n) 46 { 47 memset(b,0,sizeof(b)); 48 memset(dp,0,sizeof(dp)); 49 scanf("%d",&m); 50 for(int i =0; i<n; i++) 51 scanf("%d",&a[i]); 52 a[n] = a[n-1]+1; 53 for(int i =n-1; i>=0; i--)//倒序统计单个数在当前段出现的次数 54 { 55 if(a[i] == a[i+1]) 56 { 57 b[i] = b[i+1]+1; 58 } 59 else b[i] =1; 60 } 61 buildrmq( );//构造RMQ函数 62 int L,R,ans; 63 for(int i=1; i<=m; i++) 64 { 65 scanf("%d %d",&L,&R); 66 L = L-1;R = R-1;//题目中是从1开始计数的； 67 int temp = bi_search(L,R);//寻找数组中最左端等于a[R]的数 68 ans = b[temp] - b[R]+1; //统计和最左边相同的数出现的次数 69 if(L == temp) printf("%d ",ans);//如果这一区间中的数字相同的话，直接左边减去右边 70 else printf("%d ",max(ans,search(L,temp-1)));//否则，寻找（L,temp）中的最大数，进行比较 71 } 72 } 73 return 0; 74 }
AC代码2：线段树
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 using namespace std; 6 const int N = 1e5+10; 7 int a[N]; 8 int ans ; 9 struct node 10 { 11 int L,R,count,fre; 12 int lcount,lfre; 13 int rcount,rfre; 14 }tree[3*N]; 15 /* 16 l,r存该节点的边界。 17 count存该节点中出现最多的数字的个数，fre存该节点中出现最多的数字。 18 lcount 存该节点左端连续出现的数字的个数， lfre存该节点左端连续出现的数字。 19 rcount 存该节点右端连续出现的数字的个数， rfre存该节点右端连续出现的数字。 20 */ 21 void build(int l,int r,int v) 22 { 23 tree[v].L = l; 24 tree[v].R = r; 25 if(l == r) 26 { 27 tree[v].fre = tree[v]. rfre = tree[v].lfre = a[r]; 28 tree[v].count = tree[v].lcount = tree[v].rcount = 1; 29 return ; 30 } 31 int mid = (l+r)/2; 32 build(l,mid,v*2); 33 build(mid+1,r,v*2+1); 34 int tmpc,tmpf; 35 if(tree[v*2].count>tree[v*2+1].count) 36 { 37 tree[v].count = tree[v*2].count; 38 tree[v].fre = tree[v*2].fre; 39 } 40 else 41 { 42 tree[v].count = tree[v*2+1].count; 43 tree[v].fre = tree[v*2+1].fre; 44 } 45 tree[v].lcount = tree[v*2].lcount; 46 tree[v].rcount = tree[v*2+1].rcount; 47 if(tree[v*2].rfre == tree[v*2+1].lfre) 48 { 49 tmpc = tree[v*2].rcount +tree[v*2+1].lcount; 50 tmpf = tree[v*2].rfre; 51 if(tree[v].count<tmpc) 52 { 53 tree[v].count = tmpc; 54 tree[v].fre = tmpf; 55 } 56 if(tree[v*2].lfre == tree[v*2+1].lfre) 57 tree[v].lcount = tree[v].lcount+tree[v*2+1].lcount; 58 if(tree[v*2].rfre == tree[v*2+1].rfre) 59 tree[v].rcount= tree[v*2].rcount+tree[v].rcount; 60 } 61 tree[v].lfre = tree[v*2].lfre; 62 tree[v].rfre = tree[v*2+1].rfre; 63 } 64 void update(int x,int y,int v) 65 { 66 int mid,s1,s2; 67 if(tree[v].L == x && tree[v].R == y) 68 { 69 if(tree[v].count>ans) 70 ans = tree[v].count; 71 return ; 72 } 73 mid = (tree[v].L+tree[v].R)/2; 74 if(y<=mid) update(x,y,v*2); 75 else if(x>mid) update(x,y,v*2+1); 76 else { 77 update(x,mid,v*2); 78 update(mid+1,y,v*2+1); 79 if(tree[v*2].rfre == tree[v*2+1].lfre) 80 { 81 if(a[x] != tree[v*2].rfre) s1 = tree[v*2].rcount; 82 else s1 = mid-x+1; 83 if(a[y]!=tree[v*2+1].lfre) s2 = tree[v*2+1].lcount; 84 else s2 = y - mid; 85 if(s1+s2>ans) ans = s1+s2; 86 } 87 } 88 } 89 int main() 90 { 91 int m,n,x,y; 92 while(scanf("%d",&m) && m) 93 { 94 scanf("%d ",&n); 95 for(int i=1;i<=m;i++) 96 scanf("%d",&a[i]); 97 build(1,m,1); 98 for(int i=1;i<=n; i++) 99 { ans = 0; 100 scanf("%d %d",&x,&y); 101 if(x==y) {printf("1 ");continue;} 102 update(x,y,1); 103 printf("%d ",ans); 104 } 105 } 106 return 0; 107 }