C. DZY Loves Colors
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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?
Input
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.
Output
For each operation 2, print a line containing the answer — sum of colorfulness.
Examples
Input
3 3
1 1 2 4
1 2 3 5
2 1 3
Output
8
Input
3 4
1 1 3 4
2 1 1
2 2 2
2 3 3
Output
3
2
1
Input
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
Output
129
Note
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
解法1：线段树 打same标记，有限次之后区间缩成一个点，这是套路。
//358ms 11500kb
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 1e5+7;
namespace segmenttree{
#define lson l,m,o*2
#define rson m+1,r,o*2+1
LL sum[maxn*4], same[maxn<<2], add[maxn<<2];
void pushup(int o){
if(same[o*2] == same[o*2+1]) same[o] = same[o*2];
else same[o] = 0;
sum[o] = sum[o*2] + sum[o*2+1];
}
void pushdown(int o, int len){
if(same[o]){
sum[o*2] += add[o] * (len - (len>>1));
sum[o*2+1] += add[o] * (len >> 1);
add[o*2] += add[o];
add[o*2+1] += add[o];
add[o] = 0;
same[o*2] = same[o*2+1] = same[o];
}
}
void build(int l, int r, int o){
if(l == r){
same[o] = l; return ;
}
int m = (l + r) / 2;
build(lson);
build(rson);
pushup(o);
}
void update(int L, int R, int c, int l, int r, int o){
if(L <= l && r <= R){
if(same[o]){//只有区间颜色相同才更新
add[o] += abs(c - same[o]);
sum[o] += abs(c - same[o]) * (r - l + 1);
same[o] = c;
return ;
}
}
pushdown(o, r - l + 1);
int m = (l + r) / 2;
if(R <= m) update(L, R, c, lson);
else if(L > m) update(L, R, c, rson);
else{
update(L, m, c, lson);
update(m + 1, R, c, rson);
}
pushup(o);
}
LL query(int L, int R, int l, int r, int o){
if(L <= l && r <= R){
return sum[o];
}
pushdown(o, r - l + 1);
int m = (l + r) / 2;
if(R <= m) return query(L, R, lson);
else if(L > m) return query(L, R, rson);
else return query(L, m, lson) + query(m + 1, R, rson);
}
}
using namespace segmenttree;
int main(){
int n, m;
scanf("%d%d", &n, &m);
build(1, n, 1);
for(int i = 1; i <= m; i++){
int cmd; scanf("%d", &cmd);
if(cmd == 1){
int l, r, x; scanf("%d%d%d", &l, &r, &x); update(l, r, x, 1, n, 1);
}
else{
int l, r; scanf("%d%d", &l, &r); printf("%lld\n", query(l, r, 1, n, 1));
}
}
return 0;
}解法2：分块，维护lazy和lazyb
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6+100;
typedef long long LL;
int n, m, num, block, belong[maxn], l[maxn], r[maxn];
LL a[maxn], b[maxn], lazy[maxn], sum[maxn],lazyb[maxn];
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, lazy[i] = -1, sum[i] = 0, lazyb[i] = 0;
}
r[num] = n;
for(int i = 1; i <= n; i++) belong[i] = (i - 1) / block + 1;
}
void update(int L, int R, int v){
if(belong[L] == belong[R]){ //同一块暴力更新
if(lazy[belong[L]] != -1){
for(int i = l[belong[L]]; i <= r[belong[L]]; i++) a[i] = lazy[belong[L]];
lazy[belong[L]] = -1;
}
for(int i = L; i <= R; i++){
b[i] += abs(a[i] - v);
sum[belong[L]] += abs(a[i] - v);
a[i] = v;
}
}
else{
//开头的那不完整的一块
if(lazy[belong[L]] != -1){
for(int i = l[belong[L]]; i <= r[belong[L]]; i++) a[i] = lazy[belong[L]]; lazy[belong[L]] = -1;
}
for(int i = L; i <= r[belong[L]]; i++){
b[i] += abs(v - a[i]);
sum[belong[L]] += abs(v - a[i]);
a[i] = v;
}
//中间最多sqrt(n)块
for(int i = belong[L] + 1; i < belong[R]; i++){
if(lazy[i] != -1){
lazyb[i] += abs(lazy[i] - v);
sum[i] += abs(lazy[i] - v) * (r[i] - l[i] + 1);
lazy[i] = v;
}
else{
for(int j = l[i]; j <= r[i]; j++){
b[j] += abs(v - a[j]);
sum[i] += abs(v - a[j]);
a[j] = v;
}
lazy[i] = v;
}
}
//末尾一块，不超过sqrt(n)
if(lazy[belong[R]] != -1){
for(int i = l[belong[R]]; i <= r[belong[R]]; i++) a[i] = lazy[belong[R]];
lazy[belong[R]] = -1;
}
for(int i = l[belong[R]]; i <= R; i++){
b[i] += abs(a[i] - v);
sum[belong[R]] += abs(a[i] - v);
a[i] = v;
}
}
}
LL query(int L, int R)
{
LL ans = 0;
if(belong[L] == belong[R]){
for(int i = L; i <= R; i++) ans += b[i] + lazyb[belong[i]];
}
else{
for(int i = L; i <= r[belong[L]]; i++) ans += b[i] + lazyb[belong[L]];
for(int i = belong[L] + 1; i < belong[R]; i++) ans += sum[i];
for(int i = l[belong[R]]; i <= R; i++) ans += b[i] + lazyb[belong[R]];
}
return ans;
}
int main(){
scanf("%d%d", &n, &m);
build();
for(int i = 1; i <= n; i++) a[i] = i, b[i] = 0;
while(m--){
int cmd;
scanf("%d", &cmd);
if(cmd == 1){
int a, b; LL v;
scanf("%d%d%lld", &a, &b, &v);
update(a, b, v);
}
else{
int a, b;
scanf("%d%d", &a, &b);
printf("%lld\n", query(a, b));
}
}
return 0;
}
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