色板游戏

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  • 色板游戏

    Solution

  • 首先这道题要维护区间内的颜色种类，那么显然要用线段树。
  • 对于30种颜色，我们可以用状压来记录。

    Code

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    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #define gc getchar()
    using namespace std;
    int sc() {
        int xx = 0, ff = 1; char cch = gc;
        while(cch < '0' || cch > '9') {
            if(cch == '-') ff = -1; cch = gc;
        }
        while(cch >= '0' && cch <='9') {
            xx = (xx << 1) + (xx << 3) + (cch ^ '0'); cch = gc;
        }
        return xx * ff;
    }
    struct LST {
        int data, tag;
    }lst[100010 << 2];
    int n, m, o;
    void build(int k, int l, int r) {
        if(l == r) {
            lst[k].data = (1 << 1);
            return ;
        }
        int mid = l + r >> 1;
        build(k << 1, l, mid);
        build(k << 1 | 1, mid + 1, r);
        lst[k].data = lst[k << 1].data | lst[k << 1 | 1].data;
    }
    void pushdown(int k, int l, int r ) {
        if(lst[k].tag == 0) return ;
        lst[k << 1].tag = lst[k << 1 | 1].tag = lst[k].tag;
        lst[k << 1].data = lst[k << 1 | 1].data = (1 << lst[k].tag);
        lst[k].tag = 0;
    }
    void modify(int k, int l, int r, int x, int y, int num) {
        if(l >= x&& r <= y) {
            lst[k].tag = num;
            lst[k].data = (1 << num);
            return ;
        }
        pushdown(k, l, r);
        int mid = l + r >> 1;
        if(x <= mid) modify(k << 1, l, mid, x, y, num);
        if(y > mid) modify(k << 1 | 1, mid + 1, r, x, y, num);
        lst[k].data = lst[k << 1].data | lst[k << 1 | 1].data;
    }
    int query(int k, int l, int r, int x, int y) {
        if(l >= x && r <= y)
            return lst[k].data;
        pushdown(k, l, r);
        int mid = l + r >> 1;
        int res = 0;
        if(x <= mid) res |= query(k << 1, l, mid, x, y);
        if(y > mid) res |= query(k << 1 | 1, mid + 1, r, x, y);
        return res; 
    }
    int main() {
    //	freopen("AK.txt", "w", stdout);/s
        n = sc(); m = sc(); o = sc();
        build(1, 1, n);
        for(int i = 1; i <= o; i++) {
            char c[2];
            cin >> c[0];
            int from = sc(), to = sc();
            if(c[0] == 'C') {
                int num = sc();
                if(from > to) swap(from, to);
                modify(1, 1, n, from, to, num);
            }
            else {
                int ans = 0;
                if(from > to) swap(from, to);
                int res = query(1, 1, n, from, to);
                for(int i = 1; i <= m; i++)
                    if(res & (1 << i)) ans++;
                printf("%d\n", ans);
            }
        }
        return 0;
    }

rp++