Poj2777（线段树+标记+状态压缩）

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题目链接> http://poj.org/problem?id=2777

题目大意：给你一个长为L的板，然后给上边的每一段涂颜色，给出一系列操作对应更新区间或输出该区间有多少中颜色，本题肯定能想到线段树来解决，然后根据T<30想到用状态压缩的方法，用位运算同时也要加上lazy标记，不然很容易超时。
注意异或^的用法 x ^ (1 << (k - 1)) = 0 表示x的第k位（从 k = 1 开始）为1否则为0。
第一次写这方面的题，写的又长又丑，搞得我挖了一发就不想再看自己写的代码了，而且本题要注意所给的区间A到B可能A

#include <iostream>#include <sstream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <set>#include <vector>using namespace std;const int maxn = 100000 + 5;int flag;struct node {    int lazy;    int sum;};node tree[maxn*4];void build(int p, int l, int r){    if (l == r) {        tree[p].sum = 1;        tree[p].lazy = 1;        return;    }    int mid = (l+r) >> 1;    build(p<<1, l, mid);    build(p<<1|1, mid+1, r);    tree[p].sum = tree[p<<1].sum|tree[p<<1|1].sum;}void push_down(int p, int l, int r, int z){    tree[p].lazy = 0;//这行不能丢，丢了就WA了    node &ll = tree[p<<1];    node &rr = tree[p<<1|1];    if ((ll.lazy^(1<<(z-1))) != 0) {        ll.lazy = 1<<(z-1);        ll.sum = ll.lazy;    }    if ((rr.lazy^(1<<(z-1))) != 0) {        rr.lazy = 1<<(z-1);        rr.sum = rr.lazy;    }}void change(int p, int l, int r, int x, int y, int z){    int &s = tree[p].lazy;    if (l >= x && r <= y) {        if ((s^(1<<(z-1))) == 0) return;        else {            s = 1<<(z-1);            tree[p].sum = s;            }        return;    }    for (int i = 0; i < 31; i++) {        if ((s ^ (1<<i)) == 0) {            push_down(p, l, r, i+1);            break;        }    }    int mid = (l+r)>>1;    if (x <= mid) change(p<<1, l, mid, x, y, z);    if (y > mid) change(p<<1|1, mid+1, r, x, y, z);    tree[p].sum = tree[p<<1].sum|tree[p<<1|1].sum;}int query(int p, int l ,int r, int x, int y){    if (l >= x && r <= y) {        return tree[p].sum;    }    for (int i = 0; i < 31; i++) {        if ((tree[p].lazy ^ (1<<i)) == 0) {            push_down(p, l, r, i+1);            break;        }    }    int mid = (l+r)>>1;    int ans = 0;    if (x <= mid) ans = ans|query(p<<1, l, mid, x, y);    if (y > mid) ans = ans|query(p<<1|1, mid+1, r, x, y);    tree[p].sum = tree[p<<1].sum|tree[p<<1|1].sum;    return ans;}int main(){    //freopen("input.txt", "r", stdin);    int L, T, op;    cin >> L >> T >> op;    memset(tree, 0, sizeof(tree));    build(1, 1, L);//建树    for (int i = 0; i < op; i++) {        char ch;        int x, y, z;        scanf(" %c", &ch);        if (ch == 'C') {            scanf("%d%d%d", &x, &y, &z);            if (x > y) swap(x, y);            change(1, 1, L, x, y, z);//更新        }        else {            scanf("%d%d", &x, &y);            if (x > y) swap(x, y);            int tmp = query(1, 1, L, x, y);//询问            int ans = 0;            for (int i = 0; i <= 30; i++) {                if ((tmp&(1<<i)) ) ans++;            }            cout << ans << endl;        }    }    return 0;}

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