bzoj 2120: 数颜色

暴力,,最暴力。。。。当然,用分块做也是可以的,用带修莫队也是可以的

 1 #include<bits/stdc++.h>
 2 #define N 100005
 3 #define LL long long
 4 #define inf 0x3f3f3f3f
 5 using namespace std;
 6 inline int ra()
 7 {
 8     int x=0,f=1; char ch=getchar();
 9     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
11     return x*f;
12 }
13 int n,m,T,f[11005],a[10005],cnt;
14 map<int ,int > mp;
15 int main()
16 {
17     n=ra(); m=ra();
18     for (int i=1; i<=n; i++)
19     {
20         int x;
21         if (!mp[x=ra()])
22             mp[x]=++cnt;
23         a[i]=mp[x];
24     }
25     for (int i=1; i<=m; i++)
26     {
27         char s[3]; scanf("%s",s);
28         if (s[0]=='Q')
29         {
30             int ans=0; T++;
31             int x=ra(),y=ra();
32             for (int j=x; j<=y; j++)
33                 if (f[a[j]]!=T) 
34                     ans++,f[a[j]]=T;
35             printf("%d
",ans);
36         }
37         else
38         {
39             int pos=ra(),col=ra();
40             if (!mp[col])
41                 mp[col]=++cnt;
42             a[pos]=mp[col];
43         }
44     }
45     return 0;
46 }
 1 #include<bits/stdc++.h>
 2 #define N 1000005
 3 #define LL long long
 4 #define inf 0x3f3f3f3f
 5 using namespace std;
 6 inline int ra()
 7 {
 8     int x=0,f=1; char ch=getchar();
 9     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
10     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
11     return x*f;
12 }
13 struct node{int pos,col,pre;}w[N];
14 struct data{int l,r,id,time;}q[N];
15 int n,m,pos[N],last[N],col[N],cnt,sz,num[N],ans[N],ans1;
16 bool vis[N];
17 int cmp(data a, data b)
18 {
19     if (pos[a.l]==pos[b.l])
20     {
21         if (a.r==b.r) return a.time<b.time;
22         return a.r<b.r;
23     }
24     return pos[a.l]<pos[b.l];
25 }
26 void cal(int x)
27 {
28     if (vis[x])
29     {
30         if (!--num[col[x]]) ans1--;
31     }
32     else 
33     {
34         if (++num[col[x]]==1) ans1++;
35     }
36     vis[x]^=1;
37 }
38 void change(int x, int c)
39 {
40     if (vis[x])
41     {
42         cal(x);
43         col[x]=c;
44         cal(x);
45     }
46     else col[x]=c;
47 }
48 int main()
49 {
50     n=ra(); m=ra();
51     for (int i=1; i<=n; i++) col[i]=ra(),last[i]=col[i];
52     for (int i=1; i<=m; i++)
53     {
54         char s[3]; scanf("%s",s);
55         int x=ra(),y=ra();
56         if (s[0]=='R')
57         {
58             w[++cnt].pos=x; w[cnt].col=y; w[cnt].pre=last[x];
59             last[x]=y;
60         }
61         else
62         {
63             q[++sz].l=x; q[sz].r=y; q[sz].id=sz; q[sz].time=cnt;
64         }
65     }
66     int block=sqrt(n);
67     for (int i=1; i<=n; i++)
68         pos[i]=(i-1)*block+1;
69     sort(q+1,q+sz+1,cmp);
70     int l=1,r=1; cal(1);
71     for (int i=1; i<=m; i++)
72     {
73         for (int j=q[i-1].time+1; j<=q[i].time; j++) change(w[j].pos,w[j].col);
74         for (int j=q[i-1].time; j>q[i].time; j--) change(w[j].pos,w[j].pre);
75         while (l<q[i].l) cal(l++);
76         while (l>q[i].l) cal(--l);
77         while (r<q[i].r) cal(++r);
78         while (r>q[i].r) cal(r--);
79         ans[q[i].id]=ans1;
80     }
81     for (int i=1; i<=sz; i++) printf("%d
",ans[i]);
82     return 0;
83 }
原文地址:https://www.cnblogs.com/ccd2333/p/6482357.html