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
【分析】 单点修改+询问区间连续最长上升序列长度。 经无数次修改后依然花式RE，懵了。 谁能救救我qwqqq，求大神带。
【代码】
//hdu 3308 #include<iostream> #include<cstdio> #define fo(i,j,k) for(i=j;i<=k;i++) #define p 400000 struct node { int sum,l,r,len,lans,rans; }t[4000001]; using namespace std; int a[1000001]; int n,x,y,L,R; inline void update(node &now,node &nowl,node &nowr) { // if(nowl.l>p || nowl.r>p || nowr.l>p || nowr.r>p) return; now.sum=nowl.sum+nowr.sum; now.len=max(nowl.len,nowr.len); if(a[nowr.l]>a[nowl.r]) now.len=max(now.len,nowl.rans+nowr.lans); now.lans=nowl.lans,now.rans=nowr.rans; if(a[nowr.l]>a[nowl.r]) { if(nowl.len==nowl.r-nowl.l+1) now.lans=max(now.lans,nowl.len+nowr.lans); if(nowr.len==nowr.r-nowr.l+1) now.rans=max(now.rans,nowr.len+nowl.rans); } } inline void build(int num,int l,int r) { t[num].l=l,t[num].r=r; if(l==r) { t[num].sum=a[l]; t[num].len=t[num].lans=t[num].rans=1; return ; } // if(num*2>p) return; build(num*2,l,(l+r)/2); build(num*2+1,(l+r)/2+1,r); update(t[num],t[num*2],t[num*2+1]); } inline void change(int num,int l,int r) { if(l==r && r==x) { t[num].sum=a[l]; t[num].len=t[num].lans=t[num].rans=1; return; } // if(num*2>p) return; int mid=(l+r)/2; if(x<=mid) change(num*2,l,mid); if(x>mid) change(num*2+1,mid+1,r); update(t[num],t[num*2],t[num*2+1]); } inline node find(int L,int R,int num) { // if(num*2>p) return t[num]; node now,nowl,nowr; if(L<=t[num].l && t[num].r<=R) return t[num]; int mid=(t[num].l+t[num].r)/2+1; if(mid<=L) return find(L,R,num*2+1); else if(mid>R) return find(L,R,num*2); nowl=find(L,R,num*2); nowr=find(L,R,num*2+1); update(now,nowl,nowr); return now; } int main() { int i,j,q,T; char k[15]; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&q); fo(i,1,n) scanf("%d",&a[i]); build(1,1,n); while(q--) { scanf("%s%d%d",k,&x,&y); if(k[0]=='U') { a[x+1]=y; change(1,1,n); } else { if(x>y) swap(x,y); printf("%d\n",find(x+1,y+1,1).len); } } } return 0; } //poj 3017 Cut the S #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define M(a) memset(a,0,sizeof a) #define fo(i,j,k) for(i=j;i<=k;i++) using namespace std; const int mxn=100005; int dp[mxn],g[mxn],q[mxn],a[mxn],sum[mxn]; int n,m; inline int read() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int main() { int i,j,k=0; n=read(),m=read(); fo(i,1,n) a[i]=read(),sum[i]=sum[i-1]+a[i]; fo(i,1,n) if(sum[i]>m) break; --i; fo(j,1,i) { k=max(k,a[j]); dp[j]=k; } int h=1,t=0; fo(i,i+1,n) { while(h<=t && sum[i]-sum[q[h]-1]>m) h++; dp[i]=dp[q[h]-1]+a[q[h]]; while(h<=t && a[q[t]]<=a[i]) t--; q[++t]=i; } printf("%d\n",dp[n]); return 0; }