方法1：块状链表
PS：块状链表太强大了，也可以实现和线段树一样的lazy操作来提高效率。每个块维护自己块的信息，这个题中可以考虑每个块维护一个map来记录每个颜色在区间中出现了多少次。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <cmath> 5 #include <map> 6 using namespace std; 7 8 const int M = 400; 9 int b[M][M]; 10 int n, m, ps; 11 int lazy[M]; 12 map<int, int> mp[M]; 13 14 void down( int i ) 15 { 16 if ( lazy[i] == -1 ) return ; 17 for ( int j = 0; j < ps && i * ps + j < n; j++ ) 18 { 19 b[i][j] = lazy[i]; 20 } 21 lazy[i] = -1; 22 } 23 24 int query( int l, int r, int v ) 25 { 26 int cur = l / ps, ncur = r / ps; 27 l = l % ps, r = r % ps; 28 int res = 0; 29 for ( int i = cur + 1; i < ncur; i++ ) 30 { 31 if ( mp[i].count(v) ) 32 { 33 res += mp[i][v]; 34 } 35 } 36 if ( cur != ncur ) 37 { 38 down(cur); 39 for ( int j = l; j < ps; j++ ) 40 { 41 if ( b[cur][j] == v ) res++; 42 } 43 down(ncur); 44 for ( int j = 0; j <= r; j++ ) 45 { 46 if ( b[ncur][j] == v ) res++; 47 } 48 } 49 else 50 { 51 down(cur); 52 for ( int j = l; j <= r; j++ ) 53 { 54 if ( b[cur][j] == v ) res++; 55 } 56 } 57 return res; 58 } 59 60 void update( int l, int r, int v ) 61 { 62 int cur = l / ps, ncur = r / ps; 63 l = l % ps, r = r % ps; 64 for ( int i = cur + 1; i < ncur; i++ ) 65 { 66 mp[i].clear(); 67 mp[i][v] = ps; 68 lazy[i] = v; 69 } 70 if ( cur != ncur ) 71 { 72 down(cur); 73 for ( int j = l; j < ps; j++ ) 74 { 75 mp[cur][b[cur][j]]--; 76 b[cur][j] = v; 77 } 78 mp[cur][v] += ps - l; 79 down(ncur); 80 for ( int j = 0; j <= r; j++ ) 81 { 82 mp[ncur][b[ncur][j]]--; 83 b[ncur][j] = v; 84 } 85 mp[ncur][v] += r + 1; 86 } 87 else 88 { 89 down(cur); 90 for ( int j = l; j <= r; j++ ) 91 { 92 mp[cur][b[cur][j]]--; 93 b[cur][j] = v; 94 } 95 mp[cur][v] += r - l + 1; 96 } 97 } 98 99 int main () 100 { 101 ps = 350; 102 while ( scanf("%d%d", &n, &m) != EOF ) 103 { 104 memset( lazy, -1, sizeof(lazy) ); 105 for ( int i = 0; i < M; i++ ) mp[i].clear(); 106 for ( int i = 0; i < n; i++ ) 107 { 108 int tmp; 109 scanf("%d", &tmp); 110 b[i / ps][i % ps] = tmp; 111 mp[i / ps][tmp]++; 112 } 113 int a, l, r, z; 114 while ( m-- ) 115 { 116 scanf("%d%d%d%d", &a, &l, &r, &z); 117 if ( a == 1 ) 118 { 119 update( l, r, z ); 120 } 121 else 122 { 123 printf("%d ", query( l, r, z )); 124 } 125 } 126 } 127 return 0; 128 }
方法2：线段树+剪枝
普通的线段树是会TLE的，所以在结点上多维护两个值：maxn和minn，如果要查询的颜色大于最大或小于最小则返回0，剪枝后比方法1快了许多，此外还有几个很容易想到的剪枝，具体见代码。
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 100001; 7 int color[N]; 8 9 struct Node 10 { 11 int l, r, lazy, maxn, minn; 12 } node[N << 2]; 13 14 void pushup( int i ) 15 { 16 int lc = i << 1, rc = lc | 1; 17 node[i].maxn = max( node[lc].maxn, node[rc].maxn ); 18 node[i].minn = min( node[lc].minn, node[rc].minn ); 19 } 20 21 void build( int i, int l, int r ) 22 { 23 node[i].l = l, node[i].r = r, node[i].lazy = -1; 24 if ( l == r ) 25 { 26 node[i].maxn = node[i].minn = node[i].lazy = color[l]; 27 return ; 28 } 29 int mid = ( l + r ) >> 1; 30 build( i << 1, l, mid ); 31 build( i << 1 | 1, mid + 1, r ); 32 pushup(i); 33 } 34 35 void pushdown( int i ) 36 { 37 if ( node[i].lazy != -1 ) 38 { 39 int lc = i << 1, rc = lc | 1; 40 node[lc].lazy = node[rc].lazy = node[i].lazy; 41 node[lc].maxn = node[rc].maxn = node[i].lazy; 42 node[lc].minn = node[rc].minn = node[i].lazy; 43 node[i].lazy = -1; 44 } 45 } 46 47 void update( int i, int l, int r, int val ) 48 { 49 if ( node[i].lazy == val ) return ; 50 if ( node[i].l == l && node[i].r == r ) 51 { 52 node[i].maxn = node[i].minn = node[i].lazy = val; 53 return ; 54 } 55 pushdown(i); 56 int mid = ( node[i].l + node[i].r ) >> 1; 57 if ( r <= mid ) 58 { 59 update( i << 1, l, r, val ); 60 } 61 else if ( l > mid ) 62 { 63 update( i << 1 | 1, l, r, val ); 64 } 65 else 66 { 67 update( i << 1, l, mid, val ); 68 update( i << 1 | 1, mid + 1, r, val ); 69 } 70 pushup(i); 71 } 72 73 int query( int i, int l, int r, int val ) 74 { 75 if ( node[i].maxn < val || node[i].minn > val ) return 0; 76 if ( node[i].lazy != -1 ) 77 { 78 if ( node[i].lazy == val ) return r - l + 1; 79 else return 0; 80 } 81 int mid = ( node[i].l + node[i].r ) >> 1; 82 if ( r <= mid ) 83 { 84 return query( i << 1, l, r, val ); 85 } 86 else if ( l > mid ) 87 { 88 return query( i << 1 | 1, l, r, val ); 89 } 90 else 91 { 92 return query( i << 1, l, mid, val ) + query( i << 1 | 1, mid + 1, r, val ); 93 } 94 } 95 96 int main () 97 { 98 int n, m; 99 while ( scanf("%d%d", &n, &m) != EOF ) 100 { 101 for ( int i = 1; i <= n; i++ ) 102 { 103 scanf("%d", color + i); 104 } 105 build( 1, 1, n ); 106 int op, a, b, c; 107 while ( m-- ) 108 { 109 scanf("%d%d%d%d", &op, &a, &b, &c); 110 a++, b++; 111 if ( op == 1 ) 112 { 113 update( 1, a, b, c ); 114 } 115 else 116 { 117 printf("%d ", query( 1, a, b, c )); 118 } 119 } 120 } 121 return 0; 122 }