http://acm.hdu.edu.cn/showproblem.php?pid=3308
LCIS
Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2959 Accepted Submission(s): 1287
Problem Description
Given n integers. You have two operations: U A B: replace the Ath number by B. (index counting from 0) Q A B: output the length of the longest consecutive increasing subsequence (LCIS) in [a, b].
Input
T in the first line, indicating the case number. Each case starts with two integers n , m(0<n,m<=105). The next line has n integers(0<=val<=105). The next m lines each has an operation: U A B(0<=A,n , 0<=B=105) OR Q A B(0<=A<=B< n).
Output
For each Q, output the answer.
Sample Input
1
10 10
7 7 3 3 5 9 9 8 1 8
Q 6 6
U 3 4
Q 0 1
Q 0 5
Q 4 7
Q 3 5
Q 0 2
Q 4 6
U 6 10
Q 0 9
Sample Output
1
1
4
2
3
1
2
5
Author
shǎ崽
【题解】:
//经典线段树的应用
//难点在于Up()函数与query函数中的else
【code】:
1 /** 2 judge status: Accepted exe.time:281ms 3 exe.memory 5760k language:C++ 4 */ 5 6 #include<iostream> 7 #include<stdio.h> 8 #include<string.h> 9 #include<algorithm> 10 11 #define N 100010 12 #define lson p<<1 13 #define rson p<<1|1 14 15 using namespace std; 16 17 struct Nod 18 { 19 int l,r; //左右节点 20 int lMax,Max,rMax; //左边连续最长递增,整段连续最长递增,右边连续最长递增， 21 }node[N<<2]; 22 23 int val[N]; 24 25 void Up(int p) 26 { 27 node[p].lMax = node[lson].lMax; 28 node[p].rMax = node[rson].rMax; 29 if(val[node[lson].r]<val[node[rson].l]) //如果左边的值小于右边的 30 { 31 if(node[lson].lMax==node[lson].r-node[lson].l+1)//如果左儿子的左边最长连续值等于整个区间，则加上右儿子的左连续区间 32 { 33 node[p].lMax+=node[rson].lMax; 34 } 35 if(node[rson].rMax==node[rson].r-node[rson].l+1)//同上 36 { 37 node[p].rMax+=node[lson].rMax; 38 } 39 //下面求整段区间的最长递增 40 node[p].Max = max(node[p].rMax,node[p].lMax); 41 node[p].Max = max(node[p].Max,node[lson].Max); 42 node[p].Max = max(node[p].Max,node[rson].Max); 43 node[p].Max = max(node[p].Max,node[lson].rMax+node[rson].lMax); 44 } 45 else //否则 46 { 47 node[p].Max = max(node[lson].Max,node[rson].Max); //即为左右两个区间的较大值 48 } 49 } 50 51 void building(int l,int r,int p) //建树 52 { 53 node[p].l = l; 54 node[p].r = r; 55 if(l==r) 56 { 57 node[p].lMax=node[p].rMax=node[p].Max=1; 58 return; 59 } 60 int mid = (l+r)>>1; 61 building(l,mid,lson); 62 building(mid+1,r,rson); 63 Up(p); 64 } 65 66 void update(int id,int v,int p) 67 { 68 if(node[p].l==id&&node[p].r==id) 69 { 70 val[id] = v; 71 return; 72 } 73 int mid = (node[p].l+node[p].r)>>1; 74 if(id<=mid) update(id,v,lson); 75 else update(id,v,rson); 76 Up(p); 77 } 78 79 int Query(int l,int r,int p) 80 { 81 if(node[p].l==l&&node[p].r==r) 82 { 83 return node[p].Max; 84 } 85 int mid = (node[p].l+node[p].r)>>1; 86 if(r<=mid) return Query(l,r,lson); 87 else if(l>mid) return Query(l,r,rson); 88 else 89 { 90 int temp = max(Query(l,mid,lson),Query(mid+1,r,rson)); 91 if(val[mid]<val[mid+1]) //分开点右边大于左边的时候则比较中间加起来是否比最大值要大。 92 { 93 //min(node[lson].rMax,mid-l+1); 取最小值，因为rMax可能会大于mid-l+1； 94 temp = max(temp,min(node[lson].rMax,mid-l+1)+min(node[rson].lMax,r-mid)); 95 } 96 return temp; 97 } 98 } 99 100 int main() 101 { 102 int t; 103 int n,m; 104 scanf("%d",&t); 105 while(t--) 106 { 107 scanf("%d%d",&n,&m); 108 int i; 109 for(i=1;i<=n;i++) scanf("%d",val+i); 110 building(1,n,1); 111 while(m--) 112 { 113 char op[5]; 114 int a,b; 115 scanf("%s%d%d",op,&a,&b); 116 if(op[0]=='U') update(a+1,b,1); 117 else if(op[0]=='Q') printf("%d ",Query(a+1,b+1,1)); 118 } 119 } 120 return 0; 121 }