nowcoder wannafly 25 E：01串

E：01 串

链接

分析：

　　线段树维护转移矩阵。每个节点是一个矩阵，区间内的矩阵乘起来就是答案矩阵。矩阵乘法满足结合律，所以线段树维护。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<iostream>
 #include<cmath>
 #include<cctype>
 #include<set>
 #include<queue>
 #include<vector>
 #include<map>
 #define Root 1, n, 1
 #define lson l, mid, rt << 1
 #define rson mid + 1, r, rt << 1 | 1
 using namespace std;
 typedef long long LL;

 inline int read() {
     int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
     for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
 }

 const int N = ;
 char s[N];
 struct Node{
     int a[][];
     void Calc(int v) {
         if (v) a[][] = , a[][] = , a[][] = , a[][] = ;
         else a[][] = , a[][] = , a[][] = , a[][] = ;
     }
 }T[N << ];
 Node operator + (const Node &A,const Node &B) {
     Node C; memset(C.a, 0x3f, sizeof(C.a));
     for (int k = ; k < ; ++k)
         for (int i = ; i < ; ++i)
             for (int j = ; j < ; ++j)
                 C.a[i][j] = min(C.a[i][j], A.a[i][k] + B.a[k][j]);
     return C;
 }
 void build(int l,int r,int rt) {
     if (l == r) { T[rt].Calc(s[l] - ''); return ; }
     int mid = (l + r) >> ;
     build(lson); build(rson);
     T[rt] = T[rt << ] + T[rt <<  | ];
 }
 void update(int l,int r,int rt,int p,int v) {
     if (l == r) { T[rt].Calc(v); return ; }
     int mid = (l + r) >> ;
     if (p <= mid) update(lson, p, v);
     else update(rson, p, v);
     T[rt] = T[rt << ] + T[rt <<  | ];
 }
 Node query(int l,int r,int rt,int L,int R) {
     if (L <= l && r <= R) return T[rt];
     int mid = (l + r) >> ;
     if (R <= mid) return query(lson, L, R);
     else if (L > mid) return query(rson, L, R);
     else return query(lson, L, R) + query(rson, L, R);
 }
 int main() {
     int n = read();
     scanf("%s", s + );
     build(Root);
     Node ans;
     for (int Q = read(); Q--; ) {
         int opt = read(), x = read(), y = read();
         if (opt == ) {
             ans = query(Root, x, y);
             printf("%d\n",min(ans.a[][], ans.a[][] + ));
         }
         else update(Root, x, y);
     }
     return ;
 }