POJ 2777 Count Color（线段树区间修改+位运算）

来源：互联网发布：网络监控工程报价单编辑：程序博客网时间：2024/06/06 00:43

题目链接：点击打开链接

题意：两种操作，一个是区间修改，一个是区间查询颜色种类数。

该题因为要不断的求某个区间的颜色种类数，我们可以用位运算的并来实现。

其他的就是线段树区间修改的经典操作了。

细节参见代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<set>#include<list>#include<deque>#include<map>#include<queue>#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int mod = 1000000000 + 7;const int INF = 1000000000;const int maxn = 100000 + 10;int T,n,t,q,l,r,c,sum[maxn<<2],setv[maxn<<2];void PushUp(int o) {    sum[o] = sum[o<<1] | sum[o<<1|1];}void pushdown(int o) {    if(setv[o]) {        setv[o<<1] = setv[o<<1|1] = setv[o];        sum[o<<1] = sum[o<<1|1] = (1<<setv[o]);        setv[o] = 0;    }}void build(int l, int r, int o) {    int m = (l + r) >> 1;    setv[o] = 0;    sum[o] = (1<<1);    if(l == r) return ;    build(l, m, o<<1);    build(m+1, r, o<<1|1);}void update(int L, int R, int c, int l, int r, int o) {    int m = (l + r) >> 1;    if(L <= l && r <= R) {        sum[o] = (1 << c);        setv[o] = c;        return ;    }    pushdown(o);    if(L <= m) update(L, R, c, l, m, o<<1);    if(m < R) update(L, R, c, m+1, r, o<<1|1);    PushUp(o);}int query(int L, int R, int l, int r, int o) {    int m = (l + r) >> 1;    if(L <= l && r <= R) {        return sum[o];    }    pushdown(o);    int ans = 0;    if(L <= m) ans |= query(L, R, l, m, o<<1);    if(m < R) ans |= query(L, R, m+1, r, o<<1|1);    PushUp(o);    return ans;}char s[10];int main() {    scanf("%d%d%d",&n, &t, &q);    build(1, n, 1);    while(q--) {        scanf("%s",s);        if(s[0] == 'C') {            scanf("%d%d%d",&l, &r, &c);            update(min(l,r), max(l,r), c, 1, n, 1);        }        else {            scanf("%d%d",&l, &r);            int ans = query(min(l,r), max(l,r), 1, n, 1), cnt = 0;            for(int i=1;i<=t;i++) {                if(ans & (1<<i)) ++cnt;            }            printf("%d\n",cnt);        }    }    return 0;}

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