標籤：否則 解題思路 dex 基本 data- trap series path 檔案夾
題意：
一段區間a一開始是1、2、3、4……n這種 每次1操作能夠將[l,r]覆蓋成x 同一時候得到abs(a[i]-x)的價值 2操作查詢[l,r]的價值
思路：
線段樹 又是一道加深線段樹理解的題
操作2是簡單的求和 線段樹基本操作 痛點在操作1
用cov表示該區間的值（假設為0說明是混合區間） 用val表示該區間的價值和
那麼在更新時就不不過找到 tree[i].l==l&&tree[i].r==r 就停止了 而是繼續向下遞迴 直到一段區間的cov>0才結束
之所以這麼做是由於要將x覆蓋到之前覆蓋過的連續的區間（事實上就是將曾經混合的區間合并成一個） 這是更新val所必須的
但如此做了我們會發現val僅僅計算到了更新過的位置 無法將這次更新down下去 那麼就須要設定一個lazy標誌
lazy記錄該區間還沒有被down下去的價值
代碼：
#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 101000#define L(x) (x<<1)#define R(x) ((x<<1)|1)typedef __int64 ll;int n,m;struct node{ int l,r,cov; ll val,lazy;}tree[N*4];ll ans;int abs(int x){ if(x<0) return -x; return x;}void up(int i){ tree[i].val=tree[L(i)].val+tree[R(i)].val;}void down(int i){ if(tree[i].cov) { tree[L(i)].cov=tree[i].cov; tree[R(i)].cov=tree[i].cov; tree[i].cov=0; } if(tree[i].lazy) { tree[L(i)].lazy+=tree[i].lazy; tree[L(i)].val+=tree[i].lazy*(tree[L(i)].r-tree[L(i)].l+1); tree[R(i)].lazy+=tree[i].lazy; tree[R(i)].val+=tree[i].lazy*(tree[R(i)].r-tree[R(i)].l+1); tree[i].lazy=0; }}void init(int l,int r,int i){ tree[i].l=l; tree[i].r=r; tree[i].val=tree[i].lazy=0; if(l==r) { tree[i].cov=l; return ; } int mid=(l+r)/2; init(l,mid,L(i)); init(mid+1,r,R(i));}void solve(int i,int key){ if(tree[i].cov) { tree[i].lazy+=abs(key-tree[i].cov); tree[i].val+=(ll)(abs(key-tree[i].cov))*(tree[i].r-tree[i].l+1); tree[i].cov=key; return ; } else { down(i); solve(L(i),key); solve(R(i),key); up(i); } tree[i].cov=key;}void update(int l,int r,int i,int key){ if(tree[i].l==l&&tree[i].r==r) { solve(i,key); return ; } down(i); int mid=(tree[i].l+tree[i].r)/2; if(r<=mid) update(l,r,L(i),key); else if(l>mid) update(l,r,R(i),key); else { update(l,mid,L(i),key); update(mid+1,r,R(i),key); } up(i);}void query(int l,int r,int i){ if(tree[i].l==l&&tree[i].r==r) { ans+=tree[i].val; return ; } down(i); int mid=(tree[i].l+tree[i].r)/2; if(r<=mid) query(l,r,L(i)); else if(l>mid) query(l,r,R(i)); else { query(l,mid,L(i)); query(mid+1,r,R(i)); } up(i);}int main(){ int p,l,r,w; scanf("%d%d",&n,&m); init(1,n,1); while(m--) { scanf("%d%d%d",&p,&l,&r); if(p==1) { scanf("%d",&w); update(l,r,1,w); } else { ans=0; query(l,r,1); printf("%I64d\n",ans); } } return 0;}
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