luoguP4113 [HEOI2012]采花
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经典颜色问题推荐博文
https://www.cnblogs.com/tyner/p/11519506.html
https://www.cnblogs.com/tyner/p/11616770.html
https://www.cnblogs.com/tyner/p/11620894.html
题意
https://www.luogu.org/problem/P4113
求一段区间中超过出现两次及以上的元素种类
分析
和其他的没啥区别,维护nxt[x], 和nxt[ nxt[x] ]即可, 还是考虑移动左端点对区间答案的影响
#include<cstdio>
#include<algorithm>
using namespace std;
#define lowbit(x) (x&-x)
const int MAX = 2000000+99;
inline int read()
char ch = getchar(); int f = 1, x = 0;
while(ch<'0' || ch>'9') if(ch=='-') f = -1; ch = getchar();
while(ch>='0' && ch<='9') x = x*10+ch-'0'; ch = getchar();
return x*f;
int n,c,m;
int nxt[MAX], lst[MAX], nnxt[MAX];
int arr[MAX], t[MAX];
struct node
int l, r, id, ans;
cmd[MAX];
bool cmp1(node a, node bb) return a.l < bb.l;
bool cmp2(node a, node bb) return a.id < bb.id;
void add(int x, int k) while(x <= n) t[x] += k, x += lowbit(x);
int query(int x)
int res = 0;
while(x) res += t[x], x -= lowbit(x);
return res;
void pre()
n = read(), c = read(), m = read();
for(int i = 1; i <= n; i++) arr[i] = read();
for(int i = n; i >= 1; i--)
nxt[i] = lst[arr[i]];
lst[arr[i]] = i;
for(int i = 1; i <= n; i++) nnxt[i] = nxt[nxt[i]];
for(int i = 1; i <= c; i++) if(lst[i] && nxt[lst[i]]) add(nxt[lst[i]], 1);//颜色数为C
//分清n,m
for(int i = 1; i <= m; i++) cmd[i].l=read(), cmd[i].r=read(), cmd[i].id = i;
sort(cmd+1, cmd+1+m, cmp1);
void solve()
int pos = 1;
for(int l = 1; l <= n; l++)
while(cmd[pos].l == l)
cmd[pos].ans = query(cmd[pos].r)-query(cmd[pos].l-1);
++pos;
if(nxt[l]) add(nxt[l], -1);
if(nnxt[l]) add(nnxt[l], 1);
sort(cmd+1, cmd+1+m, cmp2);
for(int i = 1; i <= m; i++) printf("%d\\n", cmd[i].ans);
int main()
pre();
solve();
return 0;
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