Codeforces Round #254 (Div. 1)C. DZY Loves Colors(线段树经典操作/分块)

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;}