CF558E A Simple Task
Posted qihoo360
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题目链接
用权值线段树维护每个字母在([l,r])出现的次数,每次修改把每个字母在区间的出现次数记下来,然后清空这段区间,再按顺序插进去就好了。
时间复杂度(O(nlog n*26))
(好久没写正常的维护和的线段树了,这次还要打清零的标记,能一遍写过,好开森)
#include <cstdio>
#include <cstring>
int o; char ch;
inline int read(){
o = 0; ch = getchar();
while(ch < '0' || ch > '9') ch = getchar();
while(ch >= '0' && ch <= '9'){ o = o * 10 + ch - '0'; ch = getchar(); }
return o;
}
const int MAXN = 100010;
struct SegTree{
#define left (now << 1)
#define right (now << 1 | 1)
int sum[MAXN << 2], s[MAXN << 2], c[MAXN << 2];
inline void pushup(int now){
sum[now] = sum[left] + sum[right];
}
inline void pushdown(int now, int k){
if(c[now]){
sum[left] = sum[right] = s[left] = s[right] = 0;
c[left] = c[right] = 1;
c[now] = 0;
}
if(s[now]){
int len = k >> 1;
sum[left] += (k - len) * s[now];
sum[right] += len * s[now];
s[left] = s[right] = s[now];
s[now] = 0;
}
}
void update(int now, int l, int r, int wl, int wr){
if(l > wr || r < wl) return;
if(l >= wl && r <= wr){ sum[now] += (r - l + 1); ++s[now]; return; }
pushdown(now, r - l + 1);
int mid = (l + r) >> 1;
update(left, l, mid, wl, wr);
update(right, mid + 1, r, wl, wr);
pushup(now);
}
void clear(int now, int l, int r, int wl, int wr){
if(l > wr || r < wl) return;
if(l >= wl && r <= wr){ sum[now] = 0; s[now] = 0; c[now] = 1; return; }
pushdown(now, r - l + 1);
int mid = (l + r) >> 1;
clear(left, l, mid, wl, wr);
clear(right, mid + 1, r, wl, wr);
pushup(now);
}
int query(int now, int l, int r, int wl, int wr){
if(l > wr || r < wl) return 0;
if(l >= wl && r <= wr) return sum[now];
pushdown(now, r - l + 1);
int mid = (l + r) >> 1, ans = 0;
ans += query(left, l, mid, wl, wr);
ans += query(right, mid + 1, r, wl, wr);
return ans;
}
}s[27];
int n, m, A, B, C, tmp[27];
char a[MAXN];
int main(){
n = read(); m = read();
scanf("%s", a);
int len = strlen(a);
for(int i = 0; i < len; ++i)
s[a[i] - 'a'].update(1, 1, n, i + 1, i + 1);
for(int i = 1; i <= m; ++i){
A = read(); B = read(); C = read();
if(C){
for(int j = 0; j < 26; ++j)
tmp[j] = s[j].query(1, 1, n, A, B), s[j].clear(1, 1, n, A, B);
int cur = A;
for(int j = 0; j < 26; ++j)
if(tmp[j])
s[j].update(1, 1, n, cur, (cur + tmp[j]) - 1), cur += tmp[j];
}
else{
for(int j = 0; j < 26; ++j)
tmp[j] = s[j].query(1, 1, n, A, B), s[j].clear(1, 1, n, A, B);
int cur = B;
for(int j = 0; j < 26; ++j)
if(tmp[j])
s[j].update(1, 1, n, (cur - tmp[j]) + 1, cur), cur -= tmp[j];
}
}
for(int i = 1; i <= n; ++i)
for(int j = 0; j < 26; ++j)
if(s[j].query(1, 1, n, i, i)){
printf("%c", j + 'a');
break;
}
return 0;
}
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