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BIT
POJ 2481
给出n个区间，第i个用\((x_i,y_i)\)表示，定义包含：i包含j等价于\(x_i\leq x_j\land y_j\leq y_i \land\lnot(x_i=x_j\land y_i=y_j)\)，问对于每个区间，有多少个区间包含它。
BIT 做法：将区间排序，对所有i包含j使得i排在j前面，然后从前往后用BIT统计出所求。
需要注意的地方是两个区间相等的情况的处理，这时可以对原始区间离散化（相同的保存一份，并且统计出个数），在更新时一次更新所有；或者不离散化，每次统计第i个区间时，如果前一个区间和它相同，直接ans[i] = ans[i-1]。
离散化的代码：
# include <cstdio># include <cstring># include <algorithm>const int maxn = 100005;int n;int a[maxn], b[maxn], r[maxn], id[maxn];int f[maxn], fc[maxn];int c[maxn];int ans[maxn];int lb(int x) { return x & -x;}bool cmp(const int x, const int y) { if (a[x] == a[y]) return b[x] > b[y]; return a[x] < a[y];}void inc(int x, int cc) { for (; x < maxn; x += lb(x)) c[x] += cc;}int GetSum(int x) { int r; for (r = 0; x > 0; x-=lb(x)) r += c[x]; return r;}int main(){ while (scanf("%d", &n), n) { for (int i = 0; i < n; ++i) { scanf("%d%d", &a[i], &b[i]); ++a[i], ++b[i]; } for (int i = 0; i < n; ++i) r[i] = i; std::sort(r, r+n, cmp); int m = 0; id[r[0]] = 0; f[0] = r[0]; memset(fc, 0, sizeof(fc)); ++fc[0]; for (int i = 1; i < n; ++i) { int idx = r[i], pidx = r[i-1]; if (a[idx]!=a[pidx] || b[idx]!=b[pidx]) ++m, f[m] = idx; ++fc[m]; id[idx] = m; } ++m; memset(c, 0, sizeof(c)); for (int i = 0; i < m; ++i) { ans[i] = GetSum(maxn-1) - GetSum( b[ f[i] ] - 1 ); inc( b[ f[i] ], fc[i] ); } for (int i = 0; i < n; ++i) { if (i) printf(" "); printf("%d", ans[ id[i] ]); } printf("\n"); } return 0;}
没有离散化的代码：
# include <cstring># include <cstdio># include <algorithm>int n;const int maxn = 100005;int mx;int x[maxn], y[maxn], r[maxn];int c[maxn];int ans[maxn];bool cmp(const int i, const int j) { if (x[i] == x[j]) return y[j] < y[i]; return x[i] < x[j];}bool judge(int i){ if (i) { int u = r[i], v = r[i-1]; if (x[u]==x[v] && y[u]==y[v]) return true; } return false;}int lb(int x) {return x&-x;}int GetSum(int x) { int r; for (r = 0; x>0; x-=lb(x)) r+=c[x]; return r;}void inc(int x) { for (;x<=mx;x+=lb(x)) ++c[x];}int main(){ while (scanf("%d", &n), n) { mx = 1; for (int i = 0; i < n; ++i) scanf("%d%d", &x[i], &y[i]), mx = std::max(mx, ++y[i]); for (int i = 0; i < n; ++i) r[i] = i; std::sort(r, r+n, cmp); memset(c, 0, sizeof(c[0])*(mx+1)); memset(ans, 0, sizeof(ans[0])*n); for (int i = 0; i < n; ++i) { int u = r[i]; if (judge(i)) ans[u] = ans[ r[i-1] ]; else ans[u] = GetSum(mx) - GetSum(y[u]-1); inc(y[u]); } for (int i = 0; i < n; ++i) { printf(i ? " %d":"%d", ans[i]); } printf("\n"); } return 0;}
POJ 3321
先处理出dfs序列，然后对每颗子树的根直接用BIT统计即可。
vector 超时了，使用邻接表的代码：
# include <cstdio># include <cstring>const int maxn = 100005;int n,m;int c[maxn];bool a[maxn];int id[maxn];int right[maxn];int cntId;int cntEdge;int first[maxn], next[maxn*2], wor[maxn*2];void Add(int u, int v){ int head = first[u]; first[u] = ++cntEdge; next[cntEdge] = head; wor[cntEdge] = u ^ v;}void dfs(int u, int fa) { id[u] = ++cntId; for (int e = first[u]; e != -1; e = next[e]) { if (wor[e]!=(u^fa)) dfs(wor[e]^u, u); } right[ id[u] ] = cntId;}int lb(int x) {return x&-x;}void Update(int x, int cc) {for(;x<=n;x+=lb(x)) c[x]+=cc;}int GetSum(int x) { int r; for (r=0;x>0;x-=lb(x)) r+=c[x]; return r;}int main(){ while (scanf("%d", &n) != EOF) { cntEdge = 0; memset(first, -1, sizeof(first)); for (int i = 1; i < n; ++i) { int u, v; scanf("%d%d", &u, &v); Add(u, v); Add(v, u); } cntId = 0; dfs(1, -1); for (int i = 1; i <= n; ++i) c[i] = lb(i), a[i] = true; scanf("%d", &m); for (int i = 0; i < m; ++i) { char od[5]; int x; scanf("%s%d", od, &x); if (od[0] == ‘C‘) { a[x] = !a[x]; Update(id[x], a[x]?1:-1); } else { printf("%d\n", GetSum(right[ id[x] ]) - GetSum(id[x]-1)); } } } return 0;}
BIT
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