旋转子段（性质分析）
作者：互联网
60分
n^2的暴力很显然嘛........
枚举每个固定点，用个指针向区间两边扫
1 #include<cstdio>
2 #include<iostream>
3 #include<cmath>
4 #include<algorithm>
5 #include<cstring>
6 #include<string>
7 #include<map>
8 #include<vector>
9 #define MAXN 100000
10 using namespace std;
11 int n;int sum;int a[MAXN];
12 int biao[MAXN];
13 int read()
14 {
15 int x=0;char c=getchar();
16 while(c<'0'||c>'9')c=getchar();
17 while(c>='0'&&c<='9')
18 {
19 x=(x<<1)+(x<<3)+(c^48);
20 c=getchar();
21 }
22 return x;
23 }
24 int maxn;
25 int main()
26 {
27 n=read();
28 for(int i=1;i<=n;++i)
29 {
30 a[i]=read();
31 if(a[i]==i)
32 {
33 sum++;
34 biao[i]=1;
35 }
36 }
37 maxn=sum;
38 for(int i=1;i<=n;++i)
39 {
40 int point_l=i-1;int point_r=i+1;int sum_me=sum;
41 while(a[point_l]!=0&&a[point_r]!=0)
42 {
43 if(biao[point_l])sum_me--;
44 if(biao[point_r])sum_me--;
45 swap(a[point_l],a[point_r]);
46 if(a[point_l]==point_l)sum_me++;
47 if(a[point_r]==point_r)sum_me++;
48 maxn=max(sum_me,maxn);
49 swap(a[point_l],a[point_r]);
50 //printf("point_l=%d point_r=%d sum_me=%d\n",point_l,point_r,sum_me);
51 point_l--;point_r++;
52 }
53 point_l=i;point_r=i+1;sum_me=sum;
54 while(a[point_l]!=0&&a[point_r]!=0)
55 {
56 if(biao[point_l])sum_me--;
57 if(biao[point_r])sum_me--;
58 swap(a[point_l],a[point_r]);
59 if(a[point_l]==point_l)sum_me++;
60 if(a[point_r]==point_r)sum_me++;
61 maxn=max(sum_me,maxn);
62 swap(a[point_l],a[point_r]);
63 //printf("---point_l=%d point_r=%d sum_me=%d\n",point_l,point_r,sum_me);
64 point_l--;point_r++;
65 }
66 }
67 printf("%d\n",maxn);
68 }
View Code
100分n*log(n)
假设有个点位置为i权值为a[i]
那么显然它想要作出贡献只能是以（i+a[i]）/2为旋转点
那么我们发现，假设我们排序一边（以固定点为关键字），
然后就我们发现（对于同一旋转点）假设从一个区间较小->区间较大，
贡献值的变化只有旋转后，变回相应位置的点，贡献++
同时会有一部分点原来是i==a[i],这些点的贡献减去，前缀和处理
1 //性质：：区间的外围一定是可作出贡献点
2 //排序后每次可贡献点++，固定点--
3 #include<cstdio>
4 #include<iostream>
5 #include<cmath>
6 #include<algorithm>
7 #include<cstring>
8 #include<string>
9 #include<map>
10 #include<vector>
11 #define MAXN 500000
12 #define ps push_back
13 #define int long long
14 using namespace std;
15 struct node{int l,r,dian,len;}e1[MAXN],e2[MAXN];
16 int a[MAXN];
17 int sum[MAXN],biao[MAXN];
18 int n,maxn;
19 bool cmp(const node &a,const node &b)
20 {
21 return (a.dian==b.dian)?(a.len<b.len):(a.dian<b.dian);
22 }
23 signed main()
24 {
25 scanf("%lld",&n);
26 for(int i=1;i<=n;++i)
27 {
28 scanf("%lld",&a[i]);
29 sum[i]=sum[i-1];
30 if(a[i]==i)
31 {
32 biao[i]++;
33 sum[i]++;
34 }
35 }
36 int tot1=0,tot2=0;
37 for(int i=1;i<=n;++i)
38 {
39 if((i+a[i])%2==0)
40 {
41 int zh=(i+a[i])/2;
42 e1[++tot1].l=min(i,a[i]);
43 e1[tot1].r=max(i,a[i]);
44 e1[tot1].dian=zh;
45 e1[tot1].len=abs(a[i]-i)+1;
46 //printf("e[%lld].dian=%lld %lld %lld\n",tot1,e1[tot1].dian,e1[tot1].l,e1[tot1].r);
47 }
48 else
49 {
50 int zh=(i+a[i])/2;
51 e2[++tot2].l=min(i,a[i]);
52 e2[tot2].r=max(i,a[i]);
53 e2[tot2].dian=zh;
54 e2[tot2].len=abs(a[i]-i)+1;
55 }
56 }
57 sort(e1+1,e1+tot1+1,cmp);
58 sort(e2+1,e2+tot2+1,cmp);
59 /* for(int i=1;i<=tot1;i++)
60 {
61
62 }*/
63 int sum1=0;
64 for(int i=1;i<=tot1;++i)
65 {
66 //printf("i===%lld\n",i);
67 //printf("e[%lld].dian=%lld %lld %lld %lld\n",i,e1[i].dian,e1[i].l,e1[i].r,e1[i].len);
68 int l,r;
69 if(e1[i].dian==e1[i-1].dian)
70 {
71 //printf("case 1:\n");
72 if(e1[i].len==e1[i-1].len)continue;
73 l=e1[i].l;r=e1[i].r;
74 int last_l=e1[i-1].l;int last_r=e1[i-1].r;
75 if(a[l]==r)sum1++;if(a[r]==l)sum1++;
76 sum1-=sum[last_l-1]-sum[l];
77 sum1-=sum[r-1]-sum[last_r];
78 maxn=max(maxn,sum1);
79 //printf("sum1=%lld \n",sum1);
80 }
81 else
82 {
83 l=e1[i].l;r=e1[i].r;
84 sum1=0;
85 sum1+=sum[e1[i].l-1];
86 sum1+=sum[n]-sum[e1[i].r];
87 if(biao[e1[i].dian]==1)sum1++;
88 maxn=max(maxn,sum1);
89 if(l==r){continue;}
90 if(a[l]==r)sum1++;
91 if(a[r]==l)sum1++;
92 //printf("l=%lld r=%lld sum1=%lld\n",l,r,sum1);
93 maxn=max(maxn,sum1);
94 }
95
96 }
97 for(int i=1;i<=tot2;++i)
98 {
99 int l=0,r=0;
100 //printf("i=%lld\n",i);
101 // //printf("e[%lld].dian=%lld %lld %lld %lld\n",i,e2[i].dian,e2[i].l,e2[i].r,e2[i].len);
102 if(e2[i].dian==e2[i-1].dian)
103 {
104 if(e2[i].len==e2[i-1].len)continue;
105 l=e2[i].l;r=e2[i].r;
106 int last_l=e2[i-1].l,last_r=e2[i-1].r;
107 if(a[l]==r)sum1++;if(a[r]==l)sum1++;
108 // printf("sum1=%lld\n",sum,a[l],r);
109 sum1-=sum[last_l-1]-sum[l];
110 sum1-=sum[r-1]-sum[last_r];
111 maxn=max(maxn,sum1);
112 //printf("case 1sum=%lld \n",sum1);
113 }
114 else
115 {
116 l=e2[i].l;r=e2[i].r;sum1=0;
117 if(a[l]==r)sum1++;
118 if(a[r]==l)sum1++;
119 sum1+=sum[l-1];
120 sum1+=sum[n]-sum[r];
121 maxn=max(maxn,sum1);
122 // printf("l=%lld r=%lld sum1=%lld\n",l,r,sum1);
123 }
124 }
125 printf("%lld\n",maxn);
126 }
127 /*
128 8
129 2 3 4 1 5 7 8 6
130 */
View Code
标签：dian,node,子段,int,旋转,define,MAXN,include,性质 来源： https://www.cnblogs.com/Wwb123/p/11329075.html