HDU 4046 - Panda (线段树 + 单点更新)

题意

给出一个字符串,有两种操作。一种是询问[l, r]里有几个wbw,还有是修改某个点的字符。

思路

可以用一个点表示这个点的前三个是不是组成wbw,是就1,不是0.这样就可以通过线段树来处理。

如果修改了一个点,就要修改另外被影响的两个点。

代码

#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 5e4 + 10;
const int MOD = 100000007;
 
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
 
int vmax[MAXN << 2];
 
char str[MAXN];
 
void PushUp(int rt)
{
    vmax[rt] = vmax[LRT] + vmax[RRT];
}
 
void Build(int rt, int l, int r)
{
    if (l == r)
    {
        if (l >= 3)
            if (str[l] == 'w' && str[l - 1] == 'b' && str[l - 2] == 'w') vmax[rt] = 1;
            else vmax[rt] = 0;
        else vmax[rt] = 0;
        return;
    }
    int mid = MID(l, r);
    Build(LC); Build(RC);
    PushUp(rt);
}
 
void Update(int rt, int l, int r, int pos)
{
    if (l == r)
    {
        if (str[l] == 'w' && str[l - 1] == 'b' && str[l - 2] == 'w') vmax[rt] = 1;
        else vmax[rt] = 0;
        return;
    }
    int mid = MID(l, r);
    if (pos <= mid) Update(LC, pos);
    else Update(RC, pos);
    PushUp(rt);
}
 
int Query(int rt, int l, int r, int L, int R)
{
    if (L <= l && r <= R) return vmax[rt];
    int mid = MID(l, r), ans = 0;
    if (L <= mid) ans += Query(LC, L, R);
    if (R > mid) ans += Query(RC, L, R);
    return ans;
}
 
int main()
{
    //ROP;
    int T, i, j, cases = 0;
    scanf("%d", &T);
    while (T--)
    {
        printf("Case %d:\n", ++cases);
        int m, n;
        scanf("%d%d", &m, &n);
        scanf("%s", str + 1);
        Build(1, 1, m);
        while (n--)
        {
            int a, b;
            scanf("%d%d%*c", &a, &b);
            if (a == 0)
            {
                int c;
                scanf("%d", &c);
                if (c - b < 2)
                {
                    puts("0");
                    continue;
                }
                printf("%d\n", Query(1, 1, m, b + 3, c + 1));
            }
            else
            {
                b++;
                char c;
                scanf("%c", &c);
                str[b] = c;
                if (b >= 3) Update(1, 1, m, b);
                if (b >= 2 && b + 1 <= m) Update(1, 1, m, b + 1);
                if (b >= 1 && b + 2 <= m) Update(1, 1, m, b + 2);
            }
 
        }
    }
    return 0;
}

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