DZY loves colors, and he enjoys painting.
On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.
DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.
DZY wants to perform m operations, each operation can be one of the following:
- Paint all the units with numbers between l and r (both inclusive) with color x.
- Ask the sum of colorfulness of the units between l and r (both inclusive).
Can you help DZY?
The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).
Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.
If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.
If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.
For each operation 2, print a line containing the answer — sum of colorfulness.
3 3
1 1 2 4
1 2 3 5
2 1 3
8
3 4
1 1 3 4
2 1 1
2 2 2
2 3 3
3
2
1
10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10
129
In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].
After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].
After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].
So the answer to the only operation of type 2 is 8.
题意
一开始,a[i]=i,b[i]=0，然后有两个操作：
1.使得[l,r]的b[i]+=fabs(x-a[i]),a[i]=x
2.查询[l,r]的b[i]和
思路一：线段树+lazy
这个应该比较轻松吧，线段树的lazy大家应该都会，这里我就不多说了。
思路二：分块+lazy
终于感觉到卿学姐之前的分块写法的漏洞了——代码量大，虽然思路清晰，但是很容易错TAT，我debug了1个小时多。
这里的关键就是如何学习lazy技巧了。
刚开始的时候我的lazy就直接保存之前输入的val，就和线段树一样那个样子更新，但是最后我写着写着发现，代码量好像有点大啊，就果断看了一下卿学姐的代码，瞬间感觉自己好傻。
这里，我们用lastval表示之前该块里面的所有的a[i]更新以后的权值，然后lazy就表示如果是重复更新的话，那么lazy[i] += abs(val - lastval[i])即可，然后这里我们再用一个sum保存整个块的val。
当询问左右区间的时候，就直接暴力即可:ans += lazy[j] + b[j]，当询问中间区间（即块）的时候，就是ans+=sum[j]
我的代码：按照卿学姐以前的习惯写的
//看看会不会爆int!数组会不会少了一维！ //取物问题一定要小心先手胜利的条件 #include <bits/stdc++.h> using namespace std; #pragma comment(linker,"/STACK:102400000,102400000") #define LL long long #define ALL(a) a.begin(), a.end() #define pb push_back #define mk make_pair #define fi first #define se second #define haha printf("haha ") const int maxn = 1e5 + 5; LL a[maxn], b[maxn], add[maxn], sum[maxn], lazy[maxn], lastval[maxn]; int block, num, belong[maxn], l[maxn], r[maxn]; int n, m; void build(){ block = sqrt(n); num = n / block; if (n % block) num++; for (int i = 1; i <= num; i++) l[i] = (i - 1) * block + 1, r[i] = i * block; for (int i = 1; i <= n; i++) belong[i] = (i - 1) / block + 1; } ///如果lazy保存的是目前要更新的值x，那么代码的复杂度就会大大增加 ///如果lazy保存的是x减pre_lazy_val，那么代码的复杂度就会大大减少，但是会多一个空间 ///叫做lastval，表示保存之前的val void update(int x, int y, int val){ if(belong[x] == belong[y]){ if (lastval[belong[x]]){ for (int i = l[belong[x]]; i <= r[belong[x]]; i++) a[i] = lastval[belong[x]]; lastval[belong[x]] = 0; } for (int i = x; i <= y; i++){ b[i] += abs(val - a[i]); sum[belong[x]] += abs(val - a[i]); a[i] = val; } return ; } ///对x的修改 if (lastval[belong[x]]){ for (int i = l[belong[x]]; i <= r[belong[x]]; i++) a[i] = lastval[belong[x]]; lastval[belong[x]] = 0; } for (int i = x; i <= r[belong[x]]; i++){ b[i] += abs(val - a[i]); sum[belong[x]] += abs(val - a[i]); a[i] = val; } ///对y的修改 if (lastval[belong[y]]){ for (int i = l[belong[y]]; i <= r[belong[y]]; i++) a[i] = lastval[belong[y]]; lastval[belong[y]] = 0; } for (int i = l[belong[y]]; i <= y; i++){ b[i] += abs(val - a[i]); sum[belong[y]] += abs(val - a[i]); a[i] = val; } ///对块的修改 for (int i = belong[x] + 1; i < belong[y]; i++){ if (lastval[i]){ lazy[i] += abs(val - lastval[i]); sum[i] += 1LL * (r[i] - l[i] + 1) * abs(val - lastval[i]); lastval[i] = val; } else { for (int j = l[i]; j <= r[i]; j++){ b[j] += abs(val - a[j]); sum[i] += abs(val - a[j]); a[j] = val;//////////////这里增加了 } lastval[i] = val; } } } LL query(int x, int y){ LL ans = 0; if (belong[x] == belong[y]){ for (int i = x; i <= y; i++) ans += b[i] + lazy[belong[x]]; return ans; } ///对x进行操作 for (int i = x; i <= r[belong[x]]; i++) ans += b[i] + lazy[belong[x]]; ///对y进行操作 for (int i = l[belong[y]]; i <= y; i++) ans += b[i] + lazy[belong[y]]; ///对块进行操作 for (int i = belong[x] + 1; i < belong[y]; i++) ans += sum[i]; return ans; } int main(){ cin >> n >> m; for (int i = 1; i <= n; i++) a[i] = i; build(); for (int i = 1; i <= m; i++){ int ty, x, y, z; scanf("%d", &ty); if (ty == 1){ scanf("%d%d%d", &x, &y, &z); update(x, y, z); } if (ty == 2){ scanf("%d%d", &x, &y); printf("%lld ", query(x, y)); } } return 0; } /* 10 10 1 1 5 3 2 1 10 1 2 7 9 2 1 10 1 10 10 11 2 1 10 1 3 8 12 2 1 10 1 1 10 3 2 1 10 */
写完了以后我就发现自己太年轻了，果然分块还能写的更加简便啊= =。 来自：链接
#include <bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) const int maxn = 1e5 + 500; 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 n , m , unit , a[maxn] ; long long extra[maxn] , lz[maxn] , ls[maxn] , b[maxn]; int main(){ n = read() , m = read(); unit = sqrt( n ); for(int i = 0 ; i < n ; ++ i) a[i] = i + 1; while( m -- ){ int op = read() , l = read() , r = read() , x; --l; --r; if( op == 1 ){ x = read(); for(int i = l ; i <= r ; ){ int idx = i / unit; int st = idx * unit;//表示目前块的左端点，但是要注意，这里没有+1 int ed = min( n , (idx + 1) * unit );//表示目前块的右端点 if(i == st && r >= ed - 1){ if( ls[idx] ){ extra[idx] += 1LL * abs(x - ls[idx]) * (ed - st) ; lz[idx] += abs(x - ls[idx]); ls[idx] = x; } else{ for(int j = st; j < ed; ++j){ b[j] += abs(a[j] - x ); extra[idx] += abs(a[j] - x); a[j] = x; } ls[idx] = x; } i = ed; } else{///对于块的两端东西进行暴力 if(ls[idx]){ for(int j = st; j < ed; ++j) a[j] = ls[idx]; ls[idx] = 0; } extra[idx] += abs(a[i] - x); b[i] += abs(a[i] - x); a[i] = x; ++i; } } }else{ long long res = 0; for(int i = l; i <= r ;){//对于中间，全都保存在extra当中？ int idx = i / unit; int st = idx * unit; int ed = min(n, (idx + 1) * unit); if(i == st && r >= ed - 1){ res += extra[idx]; i = ed; }else{///对于两端进行暴力 res += b[i] + lz[idx]; ++i; } } printf("%lld " , res); } } return 0; }