事实上我不知道该把这道题放在那一块........
此题用了1部分二分的思想，还有函数思想，sdfz大佬用的三分思想orz.......
40分暴力：
将环搞成2len的序列，然后与处理出
sumlb[i]表示从1-i的B移动到左段的总步数 blcnt[i]表示从1-i的b的个数，其余相似。。。
然后分n个区间枚举断点，有一点：
kx=(sum_lb[k]-sum_lb[l-1])+(sum_rb[k+1]-sum_rb[r+1]);
kxl=rlcnt[l-1]*(blcnt[k]-blcnt[l-1]);
kxr=rrcnt[r+1]*(brcnt[k+1]-brcnt[r+1]);
return kx-kxl-kxr
这里我们不能单纯加kx因为数组表示从1-i而此时区间l-r则不是
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<algorithm> 6 #include<cmath> 7 #include<stack> 8 #include<map> 9 #include<queue> 10 #define ps push_back 11 #define MAXN 105101 12 #define ll long long 13 using namespace std; 14 int T; 15 int n; 16 char c[MAXN]; 17 int a[MAXN]; 18 int len; 19 int minn; 20 int sum1,sum2; 21 int sum_lb[MAXN],sum_rb[MAXN]; 22 int blcnt[MAXN],brcnt[MAXN],rrcnt[MAXN],rlcnt[MAXN]; 23 int getsum(int l,int k,int r) 24 { 25 //printf("sum_rb[%d]=%d s[%d]=%d ",k+1,sum_rb[k+1],r+1,sum_rb[r+1]); 26 int kx1=(sum_lb[k]-sum_lb[l-1])+(sum_rb[k+1]-sum_rb[r+1]); 27 int kxl=rlcnt[l-1]*(blcnt[k]-blcnt[l-1]); 28 int kxr=rrcnt[r+1]*(brcnt[k+1]-brcnt[r+1]); 29 //printf("kx1=%d lxl=%d kxr=%d ",kx1,kxl,kxr); 30 return (kx1-kxl-kxr); 31 } 32 void work(int l,int r) 33 { 34 //printf("l=%d r=%d ",l,r); 35 int ans=0; 36 for(int i=l;i<=r;++i) 37 { 38 ans=getsum(l,i,r); 39 //printf("1===i=%d ans=%d ",i,ans); 40 minn=min(ans,minn); 41 } 42 } 43 44 int main() 45 { 46 //freopen("text.in","r",stdin); 47 //freopen("wa.out","w",stdout); 48 scanf("%d",&T); 49 while(T--) 50 { 51 memset(sum_lb,0,sizeof(sum_lb)); 52 memset(sum_rb,0,sizeof(sum_rb)); 53 memset(blcnt,0,sizeof(blcnt)); 54 memset(brcnt,0,sizeof(brcnt)); 55 memset(rlcnt,0,sizeof(rlcnt)); 56 memset(rrcnt,0,sizeof(rrcnt)); 57 memset(a,0,sizeof(a)); 58 minn=1000000000; 59 //sum1=0;sum2=0; 60 scanf("%s",c+1); 61 int len=strlen(c+1); 62 if(len==1){printf("0 ");continue;} 63 for(int i=1;i<=len;++i) 64 { 65 if(c[i]=='B')a[i]=1; 66 else c[i]=2; 67 a[len+i]=a[i]; 68 } 69 int lr=0; 70 for(int i=1;i<=2*len;++i) 71 { 72 sum_lb[i]=sum_lb[i-1]; 73 blcnt[i]=blcnt[i-1];rlcnt[i]=rlcnt[i-1]; 74 if(a[i]==1) 75 { 76 sum_lb[i]+=lr; 77 blcnt[i]++; 78 //printf("sum_lb[%d]=%d ",i,sum_lb[i]); 79 } 80 else 81 { 82 lr++; 83 rlcnt[i]++; 84 } 85 //printf("bl=%d rl=%d ",blcnt[i],rlcnt[i]); 86 } 87 int rr=0; 88 for(int i=2*len;i>=1;--i) 89 { 90 sum_rb[i]=sum_rb[i+1]; 91 rrcnt[i]=rrcnt[i+1];brcnt[i]=brcnt[i+1]; 92 if(a[i]==1) 93 { 94 sum_rb[i]+=rr; 95 brcnt[i]++; 96 //printf("sum_rb[%d]=%d ",i,sum_rb[i]); 97 } 98 else 99 { 100 rr++; 101 rrcnt[i]++; 102 } 103 } 104 for(int i=1;i<=len-1;++i) 105 { 106 work(i,len+i-1); 107 } 108 printf("%d ",minn); 109 } 110 } 111 /* 112 9 113 BRRRBRRRBBRBBR 114 RRRRRBRRRRRBBRR 115 */
100分代码
满足性质区间向右移最优断点向右移或不变
看过znn的博客受到启发
然后O(n)枚举区间，指针移动最优断点即可
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<string> 5 #include<algorithm> 6 #include<cmath> 7 #include<stack> 8 #include<map> 9 #include<queue> 10 #define ps push_back 11 #define MAXN 2005101 12 #define ll long long 13 using namespace std; 14 ll T; 15 char c[MAXN]; 16 ll a[MAXN]; 17 ll len; 18 ll minn;ll zh; 19 ll sum1,sum2; 20 ll sum_lb[MAXN],sum_rb[MAXN]; 21 ll blcnt[MAXN],brcnt[MAXN],rrcnt[MAXN],rlcnt[MAXN]; 22 ll getsum(ll l,ll k,ll r) 23 { 24 ll kx1=(sum_lb[k]-sum_lb[l-1])+(sum_rb[k+1]-sum_rb[r+1]); 25 ll kxl=rlcnt[l-1]*(blcnt[k]-blcnt[l-1]); 26 ll kxr=rrcnt[r+1]*(brcnt[k+1]-brcnt[r+1]); 27 return (kx1-kxl-kxr); 28 } 29 int main() 30 { 31 scanf("%lld",&T); 32 while(T--) 33 { 34 memset(sum_lb,0,sizeof(sum_lb)); 35 memset(sum_rb,0,sizeof(sum_rb)); 36 memset(blcnt,0,sizeof(blcnt)); 37 memset(brcnt,0,sizeof(brcnt)); 38 memset(rlcnt,0,sizeof(rlcnt)); 39 memset(rrcnt,0,sizeof(rrcnt)); 40 memset(a,0,sizeof(a)); 41 minn=0x7fffffffffffffff; 42 //printf("%lld ",minn); 43 scanf("%s",c+1); 44 ll len=strlen(c+1); 45 if(len==1){printf("0 ");continue;} 46 for(ll i=1;i<=len;++i) 47 { 48 if(c[i]=='B')a[i]=1; 49 else a[i]=2; 50 a[len+i]=a[i]; 51 } 52 ll lr=0; 53 for(ll i=1;i<=2*len;++i) 54 { 55 sum_lb[i]=sum_lb[i-1]; 56 blcnt[i]=blcnt[i-1];rlcnt[i]=rlcnt[i-1]; 57 if(a[i]==1) 58 { 59 sum_lb[i]+=lr; 60 blcnt[i]++; 61 } 62 else 63 { 64 lr++; 65 rlcnt[i]++; 66 } 67 } 68 ll rr=0; 69 for(ll i=2*len;i>=1;--i) 70 { 71 sum_rb[i]=sum_rb[i+1]; 72 rrcnt[i]=rrcnt[i+1];brcnt[i]=brcnt[i+1]; 73 if(a[i]==1) 74 { 75 sum_rb[i]+=rr; 76 brcnt[i]++; 77 } 78 else 79 { 80 rr++; 81 rrcnt[i]++; 82 } 83 } 84 ll ans=0; 85 zh=(rlcnt[len]+1)/2; 86 for(ll i=1;i<=len;++i) 87 { 88 if(rlcnt[i]==zh) 89 { 90 zh=i; 91 break; 92 } 93 } 94 ans=getsum(1,zh,len); 95 //printf("aa%lld %lld ",getsum(1,zh,len),zh); 96 ans=min(ans,minn); 97 for(ll i=1;i<=len;++i) 98 { 99 ll hh=i; 100 ll tail=i+len-1; 101 if(a[i]==2) 102 { 103 ans-=blcnt[zh]-blcnt[hh-1]; 104 ans+=brcnt[zh]-brcnt[tail+1]; 105 while(a[++zh]==1) 106 { 107 ans-=rrcnt[zh]-rrcnt[tail+2]; 108 ans+=rlcnt[zh]-rlcnt[hh]; 109 } 110 //printf("ans=%lld ",ans); 111 } 112 minn=min(ans,minn); 113 } 114 //printf("%lld ",minn); 115 printf("%lld ",minn); 116 } 117 } 118 /* 119 9 120 BRRRBRRRBBRBBR 121 RRRRRBRRRRRBBRR 122 */