Probably meaning of the questions is to give you a string, 1e5 revision, each time to a range of ascending or descending order, the final output of the string;
Is actually a pretty bare tree line optimization problem; but I do not have awareness to combine barrel row, rushed to the .....
First algorithm 1.40 points
O (NMlogN)
1 inline void update(int k){ 2 for(int i=1;i<=26;++i){ 3 tmp[k].tong[i]=tmp[ls].tong[i]+tmp[rs].tong[i]; 4 } 5 } 6 7 void ud(int k,int l,int r,int L,int R){ 8 if(L<=l&&r<=R)return void(); 9 if(L<=mid)ud(ls,l,mid,L,R); 10 if(R>mid)ud(rs,mid+1,r,L,R); 11 update(k); 12
Direct sort
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define val(x) (x-'a'+1) 4 int s[100050],N,M; 5 char S[100050]; 6 bool cmp1(int a,int b){return a<b;} 7 bool cmp0(int a,int b){return a>b;} 8 int main() 9 { 10 //freopen("da.in","r",stdin); 11 // freopen("ac.out","w",stdout); 12 scanf("%d%d",&N,&M); 13 scanf("%s",S+1); 14 for(int i=1;i<=N;++i)s[i]=val(S[i]); 15 for(int i=1,l,r,opt;i<=M;++i){ 16 scanf("%d%d%d",&l,&r,&opt); 17 if(!opt)sort(s+l,s+r+1,cmp0); 18 else sort(s+l,s+r+1,cmp1); 19 // if(clock()>999900)break; 20 } 21 char c; 22 for(int i=1;i<=N;++i){ 23 c=s[i]+'a'-1; 24 printf("%c",c); 25 } 26 printf("\n"); 27 }
2. Secondly, the algorithm is still 40 points:
In fact, the topic is ungrateful people, we can directly barrels row, complexity O (NM); at least down a log, but still 40 points;
Code is not provided;
Finally 3.100 points algorithm:
Idea: combine in row 2. barrels idea, we found that one can only change the sorting O (N) complexity, and N, M <= 1e5, naturally think of tree line optimization barrels row;
And think of the complexity of the bare idea is O (MNlogN) calcium oxide! As with violence!
But you can find a range of values N is your letter, that only 26 N friends;
Finally, O (MlogN * 26)
Analyze M is asking; logN is the tree line query (merger barrels and buckets you need to query the modified region to synthesize a new bucket); 26 is required for each barrel merge complexity;
Well, then, every time we modify a range of sorts, the corresponding section of the barrel is to be merged into a new bucket, and then modify with this new range of barrels, but note that this section must not be modified to really O (N) interval assignment
It would become a real O (NMlogN) a;
How to do that?
We can play tag lazy ah! Do not say you do not like me, marking ......
Then we can mark which sections were modified and then again down access to it, the last time dfs down to a full statistical answer to it;
Note down here very long
1 void down(int k){ 2 if(!tmp[k].lazy)return ; 3 register int sz=len(ls),pl=0,pr=0; 4 memset(tmp[ls].tong,0,sizeof(tmp[ls].tong)); 5 memset(tmp[rs].tong,0,sizeof(tmp[rs].tong)); 6 if(tmp[k].opt){ 7 for(int i=1;i<=26;++i){ 8 if(tmp[k].tong[i]<=sz) 9 { 10 tmp[ls].tong[i]=tmp[k].tong[i],sz-=tmp[k].tong[i]; 11 } 12 else { 13 tmp[ls].tong[i]=sz;sz=i;break; 14 } 15 } 16 tmp[rs].tong[sz]=tmp[k].tong[sz]-tmp[ls].tong[sz]; 17 if(len(rs)==1&&tmp[rs].tong[sz]==1){pr=sz;} 18 register int zs=len(rs)-tmp[rs].tong[sz]; 19 for(int i=sz+1;i<=26;++i){ 20 tmp[rs].tong[i]=tmp[k].tong[i]; 21 zs-=tmp[k].tong[i]; 22 if(!zs)break; 23 } 24 } 25 else { 26 for(int i=26;i>=1;--i){ 27 if(tmp[k].tong[i]<=sz) 28 { 29 tmp[ls].tong[i]=tmp[k].tong[i],sz-=tmp[k].tong[i]; 30 } 31 else { 32 tmp[ls].tong[i]=sz;sz=i;break; 33 } 34 } 35 tmp[rs].tong[sz]=tmp[k].tong[sz]-tmp[ls].tong[sz]; 36 register int zs=len(rs)-tmp[rs].tong[sz]; 37 for(int i=sz-1;i>=1;--i){ 38 tmp[rs].tong[i]=tmp[k].tong[i]; 39 zs-=tmp[k].tong[i]; 40 if(!zs)break; 41 } 42 } 43 if(tmp[ls].l!=tmp[ls].r)tmp[ls].lazy=1; 44 else tmp[ls].lz=pl;tmp[ls].opt=tmp[k].opt; 45 if(tmp[rs].l!=tmp[rs].r)tmp[rs].lazy=1; 46 else tmp[rs].lz=pr;tmp[rs].opt=tmp[k].opt; 47 tmp[k].lazy=0;tmp[k].opt=0; 48 }
Of course, I'm not a Deepin the same indent;
In fact, down here, it is to segment your tree this node's two sons, the process of re-assignment barrel range corresponds, of course, once the assignment requires O (52);
Tip：
Another point is particularly prone to error
Remember that after you every time interval corresponding to the modified merger to be modified barrel barrel the corresponding section (section modification) immediately :( lg here this vector is stored in the segment tree you change the interval node number)
1 void exch(int opt){ 2 sort(lg.begin(),lg.end(),cmp); 3 for(int i=0,sz,tt=1,ts=26;i<lg.size();++i){ 4 sz=len(lg[i]); 5 memset(tmp[lg[i]].tong,0,sizeof(tmp[lg[i]].tong)); 6 if(opt){ 7 for(;tt<=26;++tt){ 8 if(sz>=ans[tt]){ 9 sz-=ans[tt]; 10 tmp[lg[i]].tong[tt]=ans[tt]; 11 } 12 else { 13 tmp[lg[i]].tong[tt]=sz; 14 ans[tt]-=sz;break; 15 } 16 } 17 } 18 else { 19 for(;ts>=1;--ts){ 20 if(sz>=ans[ts]){ 21 sz-=ans[ts]; 22 tmp[lg[i]].tong[ts]=ans[ts]; 23 } 24 else { 25 tmp[lg[i]].tong[ts]=sz; 26 ans[ts]-=sz;break; 27 } 28 } 29 } 30 } 31 }
Then immediately after you modify those changes to the root zone update again, after all, you can only maintain the lazy mark son not maintain his father;
1 inline void update(int k){ 2 for(int i=1;i<=26;++i){ 3 tmp[k].tong[i]=tmp[ls].tong[i]+tmp[rs].tong[i]; 4 } 5 } 6 7 8 void ud(int k,int l,int r,int L,int R){ 9 if(L<=l&&r<=R)return void(); 10 if(L<=mid)ud(ls,l,mid,L,R); 11 if(R>mid)ud(rs,mid+1,r,L,R); 12 update(k); 13 }
Then attach the code
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define val(x) (x-'a'+1) 4 #define mid (l+r>>1) 5 #define ls (k<<1) 6 #define rs (k<<1|1) 7 #define len(x) (tmp[x].r-tmp[x].l+1) 8 int s[100050],N,M; 9 char S[100050]; 10 struct node{ 11 int l;int r; 12 int tong[30]; 13 int lazy,lz,opt; 14 }tmp[100050<<3]; 15 int ans[30]; 16 vector<int>lg; 17 18 inline void update(int k){ 19 for(int i=1;i<=26;++i){ 20 tmp[k].tong[i]=tmp[ls].tong[i]+tmp[rs].tong[i]; 21 } 22 } 23 24 void down(int k){ 25 if(!tmp[k].lazy)return ; 26 register int sz=len(ls),pl=0,pr=0; 27 memset(tmp[ls].tong,0,sizeof(tmp[ls].tong)); 28 memset(tmp[rs].tong,0,sizeof(tmp[rs].tong)); 29 if(tmp[k].opt){ 30 for(int i=1;i<=26;++i){ 31 if(tmp[k].tong[i]<=sz) 32 { 33 tmp[ls].tong[i]=tmp[k].tong[i],sz-=tmp[k].tong[i]; 34 // if(len(ls)==1&&tmp[ls].tong[i]){pl=i;break;} 35 } 36 else { 37 tmp[ls].tong[i]=sz;sz=i;break; 38 } 39 } 40 //for(int i=1;i<=sz-1;++i)tmp[rs].tong[i]=0; 41 //for(int i=sz+1;i<=26;++i)tmp[ls].tong[i]=0; 42 tmp[rs].tong[sz]=tmp[k].tong[sz]-tmp[ls].tong[sz]; 43 if(len(rs)==1&&tmp[rs].tong[sz]==1){pr=sz;} 44 register int zs=len(rs)-tmp[rs].tong[sz]; 45 for(int i=sz+1;i<=26;++i){ 46 tmp[rs].tong[i]=tmp[k].tong[i]; 47 zs-=tmp[k].tong[i]; 48 if(!zs)break; 49 // if(len(rs)==1&&tmp[rs].tong[i]&&!pr)pr=i; 50 } 51 } 52 else { 53 for(int i=26;i>=1;--i){ 54 if(tmp[k].tong[i]<=sz) 55 { 56 tmp[ls].tong[i]=tmp[k].tong[i],sz-=tmp[k].tong[i]; 57 // if(len(ls)==1&&tmp[ls].tong[i]){pl=i;break;} 58 } 59 else { 60 tmp[ls].tong[i]=sz;sz=i;break; 61 } 62 } 63 //for(int i=26;i>=sz+1;--i)tmp[rs].tong[i]=0; 64 //for(int i=1;i<=sz-1;++i)tmp[ls].tong[i]=0; 65 tmp[rs].tong[sz]=tmp[k].tong[sz]-tmp[ls].tong[sz]; 66 //if(len(rs)==1&&tmp[rs].tong[sz]==1){pr=sz;} 67 register int zs=len(rs)-tmp[rs].tong[sz]; 68 for(int i=sz-1;i>=1;--i){ 69 tmp[rs].tong[i]=tmp[k].tong[i]; 70 zs-=tmp[k].tong[i]; 71 if(!zs)break; 72 // if(len(rs)==1&&tmp[rs].tong[i]&&!pr)pr=i; 73 } 74 } 75 if(tmp[ls].l!=tmp[ls].r)tmp[ls].lazy=1; 76 else tmp[ls].lz=pl;tmp[ls].opt=tmp[k].opt; 77 if(tmp[rs].l!=tmp[rs].r)tmp[rs].lazy=1; 78 else tmp[rs].lz=pr;tmp[rs].opt=tmp[k].opt; 79 tmp[k].lazy=0;tmp[k].opt=0; 80 } 81 82 inline void merge(int u,int opt){ 83 down(u); 84 for(register int i=1;i<=26;++i){ 85 ans[i]+=tmp[u].tong[i]; 86 } 87 if(len(u)>1)tmp[u].lazy=1;lg.push_back(u); 88 tmp[u].opt=opt; 89 } 90 91 inline int zip(int k){ 92 //cout<<"laidao!"<<endl; 93 for(int i=1;i<=26;++i)if(tmp[k].tong[i])return i; 94 } 95 96 void find(int k,int l,int r){ 97 if(l==r)return s[l]=zip(k),void(); 98 down(k); 99 find(ls,l,mid); 100 find(rs,mid+1,r); 101 } 102 103 void query(int k,int l,int r,int L,int R,int opt){ 104 if(L<=l&&r<=R)return merge(k,opt),void(); 105 down(k); 106 if(L<=mid)query(ls,l,mid,L,R,opt); 107 if(R>mid)query(rs,mid+1,r,L,R,opt); 108 } 109 110 void ud(int k,int l,int r,int L,int R){ 111 if(L<=l&&r<=R)return void(); 112 if(L<=mid)ud(ls,l,mid,L,R); 113 if(R>mid)ud(rs,mid+1,r,L,R); 114 update(k); 115 } 116 117 void building(int k,int l,int r){ 118 tmp[k].l=l;tmp[k].r=r; 119 if(l==r)return tmp[k].l=tmp[k].r=l,tmp[k].tong[s[l]]=1,tmp[k].lazy=tmp[k].opt=0,void(); 120 building(ls,l,mid);building(rs,mid+1,r); 121 update(k); 122 } 123 124 bool cmp(int u,int v){return tmp[u].l<tmp[v].l;} 125 126 void exch(int opt){ 127 sort(lg.begin(),lg.end(),cmp); 128 for(int i=0,sz,tt=1,ts=26;i<lg.size();++i){ 129 sz=len(lg[i]); 130 memset(tmp[lg[i]].tong,0,sizeof(tmp[lg[i]].tong)); 131 if(opt){ 132 for(;tt<=26;++tt){ 133 if(sz>=ans[tt]){ 134 sz-=ans[tt]; 135 tmp[lg[i]].tong[tt]=ans[tt]; 136 } 137 else { 138 tmp[lg[i]].tong[tt]=sz; 139 ans[tt]-=sz;break; 140 } 141 } 142 } 143 else { 144 for(;ts>=1;--ts){ 145 if(sz>=ans[ts]){ 146 sz-=ans[ts]; 147 tmp[lg[i]].tong[ts]=ans[ts]; 148 } 149 else { 150 tmp[lg[i]].tong[ts]=sz; 151 ans[ts]-=sz;break; 152 } 153 } 154 } 155 } 156 } 157 int main() 158 { 159 //freopen("da.in","r",stdin); 160 //freopen("te.out","w",stdout); 161 scanf("%d%d",&N,&M); 162 scanf("%s",S+1); 163 for(int i=1;i<=N;++i)s[i]=val(S[i]); 164 building(1,1,N); 165 for(int i=1,l,r,opt;i<=M;++i){ 166 scanf("%d%d%d",&l,&r,&opt); memset (years1670,sizeof(ans));lg.clear(); 168 query(1,1,N,l,r,opt); 169 exch(opt); 170 ud(1,1,N,l,r); 171 } 172 find(1,1,N); 173 char c; 174 for(int i=1;i<=N;++i){ 175 c=s[i]+'a'-1; 176 printf("%c",c); 177 } 178 printf("\n"); 179 //int ppx='P'-'a'+1; 180 //cout<<"ppx="<<ppx<<endl; 181 }