把所有询问离线一起二分回答。
跟CDQ分治比较：
CDQ分治是按照中序遍历来解决，整体二分是从上往下。
推荐一个博客。
例题洛谷P2617 Dynamic Rankings
1 #include <cstdio> 2 #include <algorithm> 3 #include <cstring> 4 5 const int N = 100010; 6 7 struct Node { 8 int f, x, y, k, id; 9 // f = 0 ask [x, y] k 10 // f = 1 change pos = x val = y sum += k 11 Node(int F = 0, int X = 0, int Y = 0, int K = 0, int ID = 0) { 12 f = F; 13 x = X; 14 y = Y; 15 k = K; 16 id = ID; 17 } 18 }node[N * 3], t1[N * 3], t2[N * 3]; 19 20 int ans[N], ta[N], a[N], n; 21 char str[3]; 22 23 inline void add(int x, int v) { 24 for(int i = x; i <= n; i += i & (-i)) { 25 ta[i] += v; 26 } 27 return; 28 } 29 30 inline int getSum(int x) { 31 int ans = 0; 32 for(int i = x; i >= 1; i -= i & (-i)) { 33 ans += ta[i]; 34 } 35 return ans; 36 } 37 38 inline void div(int L, int R, int l, int r) { 39 //printf("[%d %d] [%d %d] ", L, R, l, r); 40 if(L > R || l > r) { 41 return; 42 } 43 if(l == r) { 44 for(int i = L; i <= R; i++) { 45 if(node[i].f == 0) { 46 ans[node[i].id] = r; 47 } 48 } 49 return; 50 } 51 int mid = (l + r) >> 1; 52 int top1 = 0, top2 = 0; 53 for(int i = L; i <= R; i++) { 54 if(node[i].f == 0) { // ask 55 int t = getSum(node[i].y) - getSum(node[i].x - 1); 56 if(node[i].k <= t) { 57 t1[++top1] = node[i]; 58 } 59 else { 60 node[i].k -= t; 61 t2[++top2] = node[i]; 62 } 63 } 64 else { // change 65 if(node[i].y <= mid) { 66 add(node[i].x, node[i].k); 67 t1[++top1] = node[i]; 68 } 69 else { 70 t2[++top2] = node[i]; 71 } 72 } 73 } 74 for(int i = 1; i <= top1; i++) { 75 if(t1[i].f) { 76 add(t1[i].x, -t1[i].k); 77 } 78 } 79 memcpy(node + L, t1 + 1, top1 * sizeof(Node)); 80 memcpy(node + L + top1, t2 + 1, top2 * sizeof(Node)); 81 div(L, L + top1 - 1, l, mid); 82 div(L + top1, R, mid + 1, r); 83 return; 84 } 85 86 int main() { 87 int m, tot = 0; 88 scanf("%d%d", &n, &m); 89 memset(ans, 0x3f, sizeof(ans)); 90 int small = ans[0], large = -ans[0]; 91 for(int i = 1; i <= n; i++) { 92 scanf("%d", &a[i]); 93 node[++tot] = Node(1, i, a[i], 1, 0); 94 small = std::min(small, a[i]); 95 large = std::max(large, a[i]); 96 } 97 for(int i = 1, x, y, k; i <= m; i++) { 98 scanf("%s", str); 99 if(str[0] == 'Q') { 100 scanf("%d%d%d", &x, &y, &k); 101 node[++tot] = Node(0, x, y, k, i); 102 } 103 else { 104 scanf("%d%d", &x, &y); 105 node[++tot] = Node(1, x, a[x], -1, 0); 106 node[++tot] = Node(1, x, y, 1, i); 107 a[x] = y; 108 small = std::min(small, y); 109 large = std::max(large, y); 110 } 111 } 112 div(1, tot, small, large); 113 for(int i = 1; i <= m; i++) { 114 if(ans[i] != ans[0]) { 115 printf("%d ", ans[i]); 116 } 117 } 118 return 0; 119 }
注意这里对原来的数视作一个插入。对修改视作删除 + 插入，跟CDQ一样。用完树状数组之后要还原也跟CDQ一样。
不用合并区间所以不用归并。注意操作数不是m。
[ZJOI2013]K大数查询
当单点修改变成区间放入一个数的时候，树套树T飞......整体二分就是把树状数组的操作变成区间加，区间求和。线段树即可。
区间覆盖标记的线段树写起来繁琐至极...要注意下传tag如果把覆盖标记覆盖了，那么还要把子节点的覆盖标记传到孙子上。
50000²爆int了...
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 5 typedef long long LL; 6 const int N = 50010; 7 8 struct Node { 9 int f, x, y, id; // 0 ask 1 change 10 LL v; 11 }node[N], t1[N], t2[N]; 12 13 LL X[N]; 14 int ans[N], xx, n; 15 int tag[N * 4], zero[N * 4]; 16 LL sum[N * 4]; 17 18 inline void pushdown(int l, int r, int o) { 19 if(zero[o]) { 20 zero[o << 1] = zero[o << 1 | 1] = 1; 21 sum[o << 1] = sum[o << 1 | 1] = 0; 22 tag[o << 1] = tag[o << 1 | 1] = 0; 23 zero[o] = 0; 24 } 25 if(tag[o]) { 26 int mid = (l + r) >> 1; 27 //zero[o << 1] = zero[o << 1 | 1] = 0; 28 int ls = o << 1, rs = o << 1 | 1; 29 if(zero[ls]) { 30 zero[ls] = 0; 31 if(l < mid) { 32 zero[ls << 1] = zero[ls << 1 | 1] = 1; 33 sum[ls << 1] = sum[ls << 1 | 1] = 0; 34 tag[ls << 1] = tag[ls << 1 | 1] = 0; 35 } 36 } 37 if(zero[rs]) { 38 zero[rs] = 0; 39 if(mid + 1 < r) { 40 zero[rs << 1] = zero[rs << 1 | 1] = 1; 41 sum[rs << 1] = sum[rs << 1 | 1] = 0; 42 tag[rs << 1] = tag[rs << 1 | 1] = 0; 43 } 44 } 45 tag[o << 1] += tag[o]; 46 tag[o << 1 | 1] += tag[o]; 47 sum[o << 1] += 1ll * (mid - l + 1) * tag[o]; 48 sum[o << 1 | 1] += 1ll * (r - mid) * tag[o]; 49 tag[o] = 0; 50 } 51 return; 52 } 53 54 void add(int L, int R, int l, int r, int o) { 55 //printf("add %d %d %d %d ", L, R, l, r); 56 if(L <= l && r <= R) { 57 if(zero[o]) { 58 zero[o] = 0; 59 if(l < r) { 60 zero[o << 1] = zero[o << 1 | 1] = 1; 61 sum[o << 1] = sum[o << 1 | 1] = 0; 62 tag[o << 1] = tag[o << 1 | 1] = 0; 63 } 64 } 65 tag[o]++; 66 sum[o] += (r - l + 1); 67 //printf("sum %d += %d = %d ", o, (r - l + 1), sum[o]); 68 return; 69 } 70 pushdown(l, r, o); 71 int mid = (l + r) >> 1; 72 if(L <= mid) { 73 add(L, R, l, mid, o << 1); 74 } 75 if(mid < R) { 76 add(L, R, mid + 1, r, o << 1 | 1); 77 } 78 sum[o] = sum[o << 1] + sum[o << 1 | 1]; 79 return; 80 } 81 82 LL ask(int L, int R, int l, int r, int o) { 83 //printf("ASK : %d %d %d %d ", L, R, l, r); 84 if(L <= l && r <= R) { 85 //printf("return sum %d = %d ", o, sum[o]); 86 return sum[o]; 87 } 88 pushdown(l, r, o); 89 int mid = (l + r) >> 1; 90 LL ans = 0; 91 if(L <= mid) { 92 ans += ask(L, R, l, mid, o << 1); 93 } 94 if(mid < R) { 95 ans += ask(L, R, mid + 1, r, o << 1 | 1); 96 } 97 return ans; 98 } 99 100 inline void solve(int L, int R, int l, int r) { 101 //printf("solve : %d %d %d %d ", L, R, l, r); 102 if(L > R) { 103 return; 104 } 105 if(l == r) { 106 for(int i = L; i <= R; i++) { 107 if(node[i].f == 2) { 108 ans[node[i].id] = r; 109 } 110 } 111 return; 112 } 113 int mid = (l + r) >> 1, top1 = 0, top2 = 0; 114 for(int i = L; i <= R; i++) { 115 if(node[i].f == 2) { // ask 116 LL t = ask(node[i].x, node[i].y, 1, n, 1); 117 //printf("id = %d t = %d ", node[i].id, t); 118 if(node[i].v <= t) { 119 t2[++top2] = node[i]; 120 } 121 else { 122 node[i].v -= t; 123 t1[++top1] = node[i]; 124 } 125 } 126 else { 127 if(node[i].v > mid) { 128 add(node[i].x, node[i].y, 1, n, 1); 129 t2[++top2] = node[i]; 130 } 131 else { 132 t1[++top1] = node[i]; 133 } 134 } 135 } 136 zero[1] = 1; sum[1] = tag[1] = 0; 137 memcpy(node + L, t1 + 1, top1 * sizeof(Node)); 138 memcpy(node + L + top1, t2 + 1, top2 * sizeof(Node)); 139 solve(L, L + top1 - 1, l, mid); 140 solve(L + top1, R, mid + 1, r); 141 return; 142 } 143 144 int main() { 145 146 int m; 147 scanf("%d%d", &n, &m); 148 memset(ans, -1, sizeof(ans)); 149 for(int i = 1; i <= m; i++) { 150 scanf("%d%d%d%lld", &node[i].f, &node[i].x, &node[i].y, &node[i].v); 151 node[i].id = i; 152 if(node[i].f == 1) { 153 X[++xx] = node[i].v; 154 } 155 } 156 std::sort(X + 1, X + xx + 1); 157 xx = std::unique(X + 1, X + xx + 1) - X - 1; 158 for(int i = 1; i <= m; i++) { 159 if(node[i].f == 1) { 160 node[i].v = std::lower_bound(X + 1, X + xx + 1, node[i].v) - X; 161 } 162 } 163 solve(1, m, 1, xx); 164 for(int i = 1; i <= m; i++) { 165 if(ans[i] != ans[0]) { 166 printf("%lld ", X[ans[i]]); 167 } 168 } 169 return 0; 170 }
洛谷P1527 矩阵乘法
二维平面的时候用二维树状数组即可。
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 5 const int N = 510, M = 60010; 6 7 struct Node { 8 int f, x, y, xx, yy, k, id; /// 0 ask 1 change 9 Node(int F = 0, int X = 0, int Y = 0, int XX = 0, int YY = 0, int K = 0, int ID = 0) { 10 f = F; 11 x = X; 12 y = Y; 13 k = K; 14 xx = XX; 15 yy = YY; 16 id = ID; 17 } 18 }node[N * N + M], t1[N * N + M], t2[N * N + M]; int tot; 19 20 int X[N * N], xx, a[N][N], n; 21 int ta[N][N], ans[M]; 22 23 inline void add(int x, int y, int v) { 24 for(int i = x; i <= n; i += i & (-i)) { 25 for(int j = y; j <= n; j += j & (-j)) { 26 ta[i][j] += v; 27 } 28 } 29 return; 30 } 31 32 inline int ask(int x, int y) { 33 int ans = 0; 34 for(int i = x; i >= 1; i -= i & (-i)) { 35 for(int j = y; j >= 1; j -= j & (-j)) { 36 ans += ta[i][j]; 37 } 38 } 39 return ans; 40 } 41 42 void solve(int L, int R, int l, int r) { 43 if(L > R) { 44 return; 45 } 46 if(l == r) { 47 for(int i = L; i <= R; i++) { 48 if(node[i].f == 0) { 49 ans[node[i].id] = r; 50 } 51 } 52 return; 53 } 54 int mid = (l + r) >> 1, top1 = 0, top2 = 0; 55 for(int i = L; i <= R; i++) { 56 if(node[i].f == 0) { // ask 57 int x = node[i].x, y = node[i].y; 58 int xx = node[i].xx, yy = node[i].yy; 59 int t = ask(xx, yy) - ask(xx, y - 1) - ask(x - 1, yy) + ask(x - 1, y - 1); 60 if(node[i].k <= t) { 61 t1[++top1] = node[i]; 62 } 63 else { 64 node[i].k -= t; 65 t2[++top2] = node[i]; 66 } 67 } 68 else { // change 69 if(node[i].k <= mid) { 70 add(node[i].x, node[i].y, 1); 71 t1[++top1] = node[i]; 72 } 73 else { 74 t2[++top2] = node[i]; 75 } 76 } 77 } 78 for(int i = 1; i <= top1; i++) { 79 if(t1[i].f) { 80 add(t1[i].x, t1[i].y, -1); 81 } 82 } 83 memcpy(node + L, t1 + 1, top1 * sizeof(Node)); 84 memcpy(node + L + top1, t2 + 1, top2 * sizeof(Node)); 85 solve(L, L + top1 - 1, l, mid); 86 solve(L + top1, R, mid + 1, r); 87 return; 88 } 89 90 int main() { 91 int q; 92 scanf("%d%d", &n, &q); 93 memset(ans, 0x7f, sizeof(ans)); 94 for(int i = 1; i <= n; i++) { 95 for(int j = 1; j <= n; j++) { 96 scanf("%d", &a[i][j]); 97 node[++tot] = Node(1, i, j, 0, 0, a[i][j], 0); 98 X[tot] = a[i][j]; 99 } 100 } 101 std::sort(X + 1, X + tot + 1); 102 int xx = std::unique(X + 1, X + tot + 1) - X - 1; 103 for(int i = 1; i <= tot; i++) { 104 node[i].k = std::lower_bound(X + 1, X + xx + 1, node[i].k) - X; 105 } 106 for(int i = 1; i <= q; i++) { 107 ++tot; 108 scanf("%d%d%d%d%d", &node[tot].x, &node[tot].y, &node[tot].xx, &node[tot].yy, &node[tot].k); 109 node[tot].id = i; 110 } 111 solve(1, tot, 1, xx); 112 for(int i = 1; i <= q; i++) { 113 if(ans[i] != ans[0]) { 114 printf("%d ", X[ans[i]]); 115 } 116 } 117 return 0; 118 }