CodeForces 444 C DZY Loves Colors
题意：有n个蛋糕，起初第i个蛋糕他的颜色值为i， 现在有2个操作， 1 x y c 在[x, y]的蛋糕上都加上一层颜色值为c的蛋糕片，加了这个蛋糕片之后，会产生一个惊喜值为abs(c-a) a为上一层蛋糕片的颜色。
2 x y 求出[x, y]区间内所有蛋糕的惊喜值之和。
题解：暴力分块处理， 如果一个块被盖上了相同的颜色，那么下次这个块又被同一种颜色蛋糕片覆盖了的话就可以打一个lz标记，并且求和。
线段树也能写， 练习分块所以分块搞了半天， 然后一直wa， 重敲了一遍就AC了， 也不知道发生了啥，可能会补上线段树的写法。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout); 4 #define LL long long 5 #define ULL unsigned LL 6 #define fi first 7 #define se second 8 #define pb push_back 9 #define lson l,m,rt<<1 10 #define rson m+1,r,rt<<1|1 11 #define max3(a,b,c) max(a,max(b,c)) 12 #define min3(a,b,c) min(a,min(b,c)) 13 #define _S(X) cout << x << ' '; 14 #define __S(x) cout << x << endl; 15 typedef pair<int,int> pll; 16 const int INF = 0x3f3f3f3f; 17 const LL mod = (int)1e9+7; 18 const int N = 2e5 + 100; 19 int l[N], r[N], belong[N]; 20 LL lz[N], sum[N], v[N], a[N], tmp[N]; 21 int n, m, tot; 22 void Build(){ 23 m = sqrt(n); 24 tot = n/m; 25 if(n%m) tot++; 26 for(int i = 1; i <= n; i++) 27 belong[i] = (i-1)/m + 1, a[i] = i; 28 for(int i = 1; i <= tot; i++) 29 l[i] = (i-1)*m+1, r[i] = i*m; 30 r[tot] = n; 31 } 32 void PushDown(int rt){ 33 for(int i = l[rt]; i <= r[rt]; i++) 34 a[i] = lz[rt]; 35 lz[rt] = 0; 36 } 37 void Update(int x, int y, LL c){ 38 int idx = belong[x], idy = belong[y]; 39 if(idx == idy){ 40 if(lz[idx]) PushDown(idx); 41 for(int i = x; i <= y; i++){ 42 sum[idx] += abs(c - a[i]); 43 v[i] += abs(c - a[i]); 44 a[i] = c; 45 } 46 } 47 else { 48 if(lz[idx]) PushDown(idx); 49 if(lz[idy]) PushDown(idy); 50 for(int i = x; i <= r[idx]; i++){ 51 sum[idx] += abs(c - a[i]); 52 v[i] += abs(c - a[i]); 53 a[i] = c; 54 } 55 for(int i = l[idy]; i <= y; i++){ 56 sum[idy] += abs(c - a[i]); 57 v[i] += abs(c - a[i]); 58 a[i] = c; 59 } 60 for(int i = idx+1; i < idy; i++){ 61 if(lz[i]){ 62 tmp[i] += abs(lz[i] - c); 63 sum[i] += (r[i]-l[i]+1)*(abs(lz[i]-c)); 64 lz[i] = c; 65 } 66 else { 67 lz[i] = c; 68 for(int j = l[i]; j <= r[i]; j++){ 69 v[j] += abs(a[j] - c); 70 sum[i] += abs(a[j]-c); 71 a[j] = c; 72 } 73 } 74 } 75 } 76 } 77 void Query(int x, int y){ 78 int idx = belong[x], idy = belong[y]; 79 LL ans = 0; 80 if(idx == idy){ 81 for(int i = x; i <= y; i++) 82 ans += (tmp[idx] + v[i]); 83 } 84 else { 85 for(int i = x; i <= r[idx]; i++) 86 ans += (tmp[idx] + v[i]); 87 for(int i = l[idy]; i <= y; i++) 88 ans += (tmp[idy] + v[i]); 89 for(int i = idx+1; i < idy; i++) 90 ans += sum[i]; 91 } 92 printf("%lld\n", ans); 93 } 94 int main(){ 95 int q, x, y, c, op; 96 scanf("%d%d", &n, &q); 97 Build(); 98 while(q--){ 99 scanf("%d%d%d", &op, &x, &y); 100 if(op == 1) { 101 scanf("%d", &c); 102 Update(x,y,c); 103 } 104 else { 105 Query(x,y); 106 } 107 108 } 109 return 0; 110 }
