题目链接https://loj.ac/problem/6284
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 typedef long long ll; 5 const int maxn = 1e5 + 10; 6 const int inf = 0x3f3f3f3f; 7 int dir[10][10] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}}; 8 int pos[maxn], l[maxn], r[maxn], v[maxn], lz[maxn]; // lz数组判定整个块是否都为同一个数，不是-1就是区间都为lz的值。 9 int a, b, c, t, n; 10 11 inline int read() 12 { 13 char ch = getchar(); int k = 0, f = 1; 14 while(ch < '0' || ch > '9') {if(ch == '-') f = -1; ch = getchar();} 15 while(ch >= '0' && ch <= '9') {k = (k << 1) + (k << 3) + ch - '0'; ch = getchar();} 16 return k * f; 17 } 18 19 inline void reset(int x) 20 { 21 if(lz[x] == -1) return ; 22 for(int i = l[x]; i <= r[x]; ++i) v[i] = lz[x]; //lz非-1的话说明此块里都是lz的值 23 lz[x] = -1; 24 } 25 inline int findy(int a,int b,int c) 26 { 27 int pa = pos[a], pb = pos[b], ans = 0; 28 if(pa == pb) 29 { 30 reset(pa); 31 for(int i = a; i <= b; ++i) 32 { 33 if(v[i] == c) ans++; v[i] = c; 34 } 35 lz[pa] = -1; // 进行了修改就不能保证lz块里一定是同一个数了 36 } 37 else 38 { 39 reset(pa); 40 for(int i = a; i <= r[pa]; ++i) {if(v[i] == c) ans++; v[i] = c;} 41 reset(pb); 42 for(int i = l[pb]; i <= b; ++i) {if(v[i] == c) ans++; v[i] = c;} 43 for(int i = pa + 1; i <= pb - 1; ++i) 44 { 45 int k = 0; 46 if(lz[i] != -1) 47 { 48 if(lz[i] == c) ans += (r[i] - l[i] + 1); 49 else lz[i] = c; 50 } 51 else 52 { 53 for(int j = l[i]; j <= r[i]; ++j) 54 { 55 if(v[j] == c) ans++; v[j] = c; 56 } 57 lz[i] = c; 58 } 59 } 60 } 61 return ans; 62 } 63 int main() 64 { 65 n = read(); 66 for(int i = 1; i <= n; ++i) 67 { 68 v[i] = read(); 69 } 70 t = sqrt(n); 71 for(int i = 1; i <= t; ++i) 72 { 73 l[i] = (i-1) * t + 1; 74 r[i] = i * t; 75 } 76 if(r[t] != n) t++, l[t] = r[t - 1] + 1, r[t] = n; 77 for(int i = 1; i <= t; ++i) 78 { 79 for(int j = l[i]; j <= r[i]; ++j) 80 { 81 pos[j] = i; 82 } 83 } 84 memset(lz, -1, sizeof(lz)); 85 for(int i = 1; i <= n; ++i) 86 { 87 a = read(); b = read(); c = read(); 88 cout<<findy(a, b, c)<<endl; 89 } 90 return 0; 91 }