Codeforces 1304F2 Animal Observation (hard version) 代码（dp滑动窗口线段树区间更新优化）
https://codeforces.com/contest/1304/problem/F2
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int maxn = 4e4+5; 4 int dp[55][maxn]; 5 int val[55][maxn]; 6 int sum[55][maxn]; 7 int n,m,k; 8 struct node{ 9 int l,r; 10 int Max,lz; 11 }seg_t[maxn*4]; 12 void build(int l,int r,int p){ 13 seg_t[p].l = l,seg_t[p].r = r; 14 if(l == r) { 15 seg_t[p].Max = 0;return ; 16 } 17 int mid = (l+r)>>1; 18 build(l,mid,p*2); 19 build(mid+1,r,p*2+1); 20 seg_t[p].Max = max(seg_t[p*2].Max ,seg_t[p*2+1].Max ); 21 } 22 void pushdown(int k){ 23 seg_t[k*2].lz +=seg_t[k].lz ; 24 seg_t[k*2+1].lz +=seg_t[k].lz ; 25 seg_t[k*2].Max +=seg_t[k].lz ; 26 seg_t[k*2+1].Max +=seg_t[k].lz ; 27 seg_t[k].lz = 0; 28 } 29 void upd(int L,int R,int p,int v){ 30 if(seg_t[p].l == L && seg_t[p].r == R){ 31 seg_t[p].lz +=v; 32 seg_t[p].Max +=v; 33 return; 34 } 35 if(seg_t[p].lz ) pushdown(p); 36 int mid = (seg_t[p].l + seg_t[p].r )>>1; 37 if(R<=mid) upd(L,R,p*2,v); 38 else if(L>mid) upd(L,R,p*2+1,v); 39 else{ 40 upd(L,mid,p*2,v); 41 upd(mid+1,R,p*2+1,v); 42 } 43 seg_t[p].Max = max(seg_t[p*2].Max ,seg_t[p*2+1].Max ); 44 } 45 int query(int p,int L,int R){ 46 if(seg_t[p].l == L && seg_t[p].r == R) return seg_t[p].Max ; 47 if(seg_t[p].lz ) pushdown(p); 48 int mid = (seg_t[p].l +seg_t[p].r)>>1; 49 if(R<=mid) return query(p*2,L,R); 50 else if(L>mid) return query(p*2+1,L,R); 51 else return max(query(p*2,L,mid),query(p*2+1,mid+1,R)); 52 } 53 int main(){ 54 cin>>n>>m>>k; 55 for(int i = 1;i<=n;i++){ 56 for(int j = 1;j<=m;j++) { 57 cin>>val[i][j]; 58 sum[i][j] = sum[i][j-1] + val[i][j]; 59 } 60 } 61 for(int i = 1;i<=m-k+1;i++){ 62 dp[1][i] = sum[1][i+k-1] - sum[1][i-1]+sum[2][i+k-1]-sum[2][i-1]; 63 } 64 for(int i = 2;i<=n;i++){ 65 memset(seg_t,0,sizeof(seg_t)); 66 build(1,2*m,1); 67 for(int j = 1;j<=m;j++) upd(j,j,1,dp[i-1][j]); 68 for(int j = 1;j<=k;j++) upd(1,j,1,-val[i][j]); 69 for(int j = 1;j<=m-k+1;j++){ 70 dp[i][j] = max(dp[i][j],query(1,1,m)+sum[i][j+k-1]-sum[i][j-1]+sum[i+1][j+k-1]-sum[i+1][j-1]); 71 upd(max(1,j-k+1),j,1,val[i][j]); 72 upd(j+1,j+k,1,-val[i][j+k]); 73 } 74 } 75 int ans = 0; 76 for(int i = 1;i<=m;i++) ans = max(ans,dp[n][i]); 77 cout<<ans; 78 return 0; 79 }
