离线处理+树状数组
//http://blog.csdn.net/zz_1215/ #pragma comment(linker, "/STACK:102400000,102400000") #include<iostream> #include<vector> #include<algorithm> #include<cstdio> #include<queue> #include<stack> #include<string> #include<map> #include<set> #include<cmath> #include<cassert> #include<cstring> #include<iomanip> #include<ctime> using namespace std; typedef long long i64; #define S64(a) scanf(in64,&a) #define SS(a) scanf("%d",&a) #define LL(a) ((a)<<1) #define RR(a) (((a)<<1)+1) #define pb push_back #define pf push_front #define X first #define Y second #define CL(Q) while(!Q.empty())Q.pop() #define MM(name,what) memset(name,what,sizeof(name)) #define MC(a,b) memcpy(a,b,sizeof(b)) #define MAX(a,b) ((a)>(b)?(a):(b)) #define MIN(a,b) ((a)<(b)?
(a):(b)) #define read freopen("in.txt","r",stdin) #define write freopen("out.txt","w",stdout) const int inf = 0x3f3f3f3f; const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL; const double oo = 10e9; const double eps = 10e-9; const double pi = acos(-1.0); const int maxn = 100000 + 111; struct zz{ int l, r, id; bool operator < (const zz& cmp) const{ return l < cmp.l; } }; int n, m; int a[maxn]; bool have[maxn]; int pos[maxn]; int ta[maxn]; int ans[maxn]; vector<zz>v; int lowbit(int x){ return x & (-x); } void add(int pos, int num, int ary[]){ while (pos <= n){ ary[pos] += num; pos += lowbit(pos); } } int sum(int x, int ary[]){ // 1 to x int ans = 0; while (x){ ans += ary[x]; x -= lowbit(x); } return ans; } int query(int x, int y){ return sum(y, ta) - sum(x - 1, ta); } inline int find(int x){ if (!have[x - 1] && !have[x + 1]){ return 1; } else if (have[x - 1] && have[x + 1]){ return -1; } else{ return 0; } } void start(){ for (int i = 1; i <= n; i++){ have[i] = false; ta[i] = 0; } for (int i = 1; i <= n; i++){ pos[a[i]] = i; } for (int i = 1; i <= n; i++){ add(i, find(a[i]), ta); have[a[i]] = true; } int temp = 0; for (int i = 1; i <= n; i++){ while (temp <(int) v.size()){ if (v[temp].l == i){ ans[v[temp].id] = query(v[temp].l, v[temp].r); temp++; } else{ break; } } have[a[i]] = false; if (have[a[i] + 1]){ add(pos[a[i] + 1], 1, ta); } if (have[a[i] - 1]){ add(pos[a[i] - 1], 1, ta); } } return; } int main() { int T; cin >> T; while (T--){ cin >> n >> m; for (int i = 1; i <= n; i++){ cin >> a[i]; } zz temp; v.clear(); for (int i = 1; i <= m; i++){ //cin >> temp.l >> temp.r; SS(temp.l); SS(temp.r); temp.id = i; v.push_back(temp); } sort(v.begin(), v.end()); start(); for (int i = 1; i <= m; i++){ //cout << ans[i] << endl; printf("%d\n", ans[i]); } } return 0; }