51nod 小朋友的笑话

链接

分析：

　　每次操作把以前没有出现这个数的设为1，有这个数的设为0。首先将当前区间设为1，考虑有set维护这个颜色出现的区间，然后把所有与当前区间相交的拿出来，修改为0。

　　复杂度？每次操作的线段只会加入到一次set中，从set中取出一次，只会修改一次，然后就合并成大的了，每次操作也只会加入一条线段。

代码：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<bitset>
#define pa pair<int,int>
using namespace std;
typedef long long LL;

inline int read() {
    int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
    for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
}

const int N = ;
set< pa > s[N];
int n;

struct SegmentTree{
    int sum[N << ], tag[N << ], n;
    SegmentTree() { memset(tag, -, sizeof(tag)); }
    void pushdown(int rt,int len) {
        tag[rt << ] = tag[rt], tag[rt <<  | ] = tag[rt];
        sum[rt << ] = tag[rt] * (len - len / );
        sum[rt <<  | ] = tag[rt] * (len / );
        tag[rt] = -;
    }
    void update(int l,int r,int rt,int L,int R,int v) {
        if (L > R) return ;
        if (L <= l && r <= R) { tag[rt] = v, sum[rt] = v * (r - l + ); return ; }
        if (~tag[rt]) pushdown(rt, r - l + );
        int mid = (l + r) >> ;
        if (L <= mid) update(l, mid, rt << , L, R, v);
        if (R > mid) update(mid + , r, rt <<  | , L, R, v);
        sum[rt] = sum[rt << ] + sum[rt <<  | ];
    }
    int query(int l,int r,int rt,int L,int R) {
        if (L <= l && r <= R) return sum[rt];
        if (~tag[rt]) pushdown(rt, r - l + );
        int mid = (l + r) >> , res = ;
        if (L <= mid) res = query(l, mid, rt << , L, R);
        if (R > mid) res += query(mid + , r, rt <<  | , L, R);
        return res;
    }
}T;
void Insert() {
    int x = read(), y = read(), k = read(), l = max(, x - k), r = min(n, x + k), nowl = l, nowr = r;
    if (s[y].empty()) {
        T.update(, n, , l, r, );
        s[y].insert(pa(l, r)); return ;
    }
    set< pa > :: iterator it = s[y].lower_bound(pa(l, ));
    set< pa > :: iterator pre = it;

    if (it != s[y].begin()) {
        pre --;
        if (pre->second >= l) it = pre;
    }
    if (it == s[y].end() || (it->first) > r) {
        T.update(, n, , l, r, );
        s[y].insert(pa(l, r)); return ;
    }
    nowl = min(nowl, it->first);
    T.update(, n, , l, r, );
    while (it != s[y].end() && (it->first) <= r) {
        T.update(, n, , max(l, it->first), min(r, it->second), );
        nowr = max(nowr, it->second);
        pre = it; it ++; s[y].erase(pre);
    }
    s[y].insert(pa(nowl, nowr));
}
void query() {
    int l = read(), r = read();
    printf("%d\n", T.query(, n, , l, r));
}
int main() {
    n = read();int m = read();
    for (int i = ; i <= m; ++i) {
        if (read() == ) Insert();
        else query();
    }
    return ;
}