POJ-2892（树状数组 + 二分）

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题目：http://poj.org/problem?id=2892

poj上1A，但是拿到hdoj1540上却WA了，看了discuss发现在hdoj需要考虑一个村庄被摧毁和恢复多次的情况

#include <cstdio>#include <cstring>#define MAX50005int N, M;int c[MAX] = {0}, b[MAX];bool a[MAX] = {0};int stack[MAX], top;inline int lowbit(int x){ return x & -x; }void update(int x, int v){if(v == 1 && a[x] || v == -1 && !a[x]) return;a[x] = (v == 1);for(; x <= N; x += lowbit(x)) c[x] += v;}int sum(int x){int res = 0;for(; x; x -= lowbit(x)) res += c[x];return res;}int upper(int v)//find leftest i that sum(i) > v{int l = 0, r = N;while(l + 1 < r){int m = (l + r) >> 1;if(sum(m) > v) r = m;else l = m;}return r;}int lower(int v)//find leftest i that sum(i) >= v{int l = 0, r = N;while(l + 1 < r){int m = (l + r) >> 1;if(sum(m) >= v) r = m;else l = m;}return r;}int query(int x){if(a[x]) return 0;int v = sum(x), r = x < N ? upper(v) : x, l = x > 1 ? lower(v) : 1;if(a[l] && a[r]) return r - l - 1;else if(a[l] || a[r]) return r - l;else return r - l + 1;}int main(){char s[4];int x;while(~scanf("%d%d", &N, &M)){top = 0;memset(a + 1, 0, N);memset(c + 1, 0, N * sizeof(int));for(int i = 0; i <= N; ++i) b[i] = i;while(M--){scanf("%s", s);if(s[0] == 'D'){scanf("%d", &x);update(x, 1);stack[top++] = x;}else if(s[0] == 'Q'){scanf("%d", &x);printf("%d\n", query(x));}else update(stack[--top], -1);}}return 0;}

时间复杂度O(n(logn)^2)，常数小的树状数组真是厉害

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