题目链接：uva 11402 - Ahoy, Pirates!

题目大意：给定给一个字符串，字符串的给定方式为各个循坏单位的循环次数和循环单位，然后是Q次操作。

• F：将l~r之间的数变成1
• E：将l~r之间的束变成0
• I：将l~r之间的数0变1，1变0
• Q：查询l~r之间1的个数

解题思路：线段树，注意pushdown函数中I操作不属于覆盖操作，要与子节点中的懒惰标记判断关系处理。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1024005;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

int N;
bool v[maxn];

struct Node {
    int l, r;
    int vset, vmix, val;
    void set (int l, int r, int val, int vset) {
        this->l = l;
        this->r = r;
        this->val = val;
        this->vset = vset;
    }
}node[maxn*4];

void set_node (int u, int vset) {
    if (vset == 1 || (node[u].vset == 2 && vset == 3)) {
        node[u].val = node[u].r - node[u].l + 1;
        node[u].vset = 1;
    } else if (vset == 2 || (node[u].vset == 1 && vset == 3)) {
        node[u].val = 0;
        node[u].vset = 2;
    } else {
        node[u].val = node[u].r - node[u].l + 1 - node[u].val;
        node[u].vset = vset - node[u].vset;
    }
}

void pushup (int u) {
    node[u].set(node[lson(u)].l, node[rson(u)].r, node[lson(u)].val + node[rson(u)].val, 0);
}

void pushdown (int u) {
    if (node[u].vset) {
        set_node(lson(u), node[u].vset);
        set_node(rson(u), node[u].vset);
    }
    node[u].vset = 0;
}

void build_segTree (int u, int l, int r) {
    if (l == r) {
        node[u].set(l, r, v[l] ? 1 : 0, 0);
        return;
    }

    int mid = (l + r) / 2;
    build_segTree(lson(u), l, mid);
    build_segTree(rson(u), mid+1, r);
    pushup(u);
}

void init () {
    N = 0;
    memset(v, 0, sizeof(v));

    int n, x;
    char s[100];
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d%s", &x, s);
        int len = strlen(s);
        for (int j = 0; j < len; j++) {
            if (s[j] == ‘1‘) {
                for (int k = 0; k < x; k++)
                    v[k*len+j+N] = 1;
            }
        }
        N += x * len;
    }
    build_segTree(1, 0, N-1);
}

void set_segTree (int u, int l, int r, int vset) {

    if (l <= node[u].l && node[u].r <= r) {
        set_node(u, vset);
        return;
    }

    pushdown(u);
    int mid = (node[u].l + node[u].r) / 2;
    if (l <= mid)
        set_segTree(lson(u), l, r, vset);
    if (r > mid)
        set_segTree(rson(u), l, r, vset);
    pushup(u);
}

int query_segTree (int u, int l, int r) {
    if (l <= node[u].l && node[u].r <= r)
        return node[u].val;

    pushdown(u);
    int mid = (node[u].l + node[u].r) / 2;
    int ret = 0;
    if (l <= mid)
        ret += query_segTree(lson(u), l, r);
    if (r > mid)
        ret += query_segTree(rson(u), l, r);
    pushup(u);
    return ret;
}

void solve () {
    int n, cas = 1, l, r;
    char s[5];

    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%s%d%d", s, &l, &r);
        if (s[0] == ‘F‘)
            set_segTree(1, l, r, 1);
        else if (s[0] == ‘E‘)
            set_segTree(1, l, r, 2);
        else if (s[0] == ‘I‘)
            set_segTree(1, l, r, 3);
        else
            printf("Q%d: %d\n", cas++, query_segTree(1, l, r));
    }
}

int main () {
    int cas;
    scanf("%d", &cas);
    for (int kcas = 1; kcas <= cas; kcas++) {
        init();

        printf("Case %d:\n", kcas);
        solve();
    }
    return 0;
}

uva 11402 - Ahoy, Pirates!(线段树)