线段树 fzu2187 回家种地

来源：互联网发布：sql union all 效率编辑：程序博客网时间：2024/05/01 13:58

一道很经典的面积并问题。

和普通的面积并相比，这个的面积并求的是只覆盖了一次的答案。

相对于普通的来说，就是求cnt==1的长度，而不是求cnt>=1的长度而已

我的想法是用S1维护cnt>=1的长度，S2维护cnt==1的长度

其他地方和普通的面积并一模一样，只是在push_up的时候稍微多了对S2的维护而已

#include<map>#include<set>#include<cmath>#include<queue>#include<cstdio>#include<string>#include<vector>#include<cstring>#include<iostream>#include<algorithm>#include<functional>using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 2e5 + 5;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define root 1,rear,1int rear, cnt[MX << 2];int S1[MX << 2], S2[MX << 2], A[MX];struct Que {    int sign;    int top, L, R;    bool operator<(const Que &b)const {        return top < b.top;    }    Que() {}    Que(int _top, int _L, int _R, int _sign) {        top = _top; L = _L; R = _R; sign = _sign;    }} Q[MX];int BS(int x) {    int l = 1, r = rear, m;    while(l <= r) {        m = (l + r) >> 1;        if(A[m] == x) return m;        if(A[m] < x)  l = m + 1;        else        r = m - 1;    }    return -1;}void push_up(int l, int r, int rt) {    if(cnt[rt]) {        S1[rt] = A[r + 1] - A[l];        if(cnt[rt] == 1) S2[rt] = S1[rt] - S1[rt << 1] - S1[rt << 1 | 1];        else S2[rt] = 0;    } else if(l == r) S1[rt] = S2[rt] = 0;    else {        S1[rt] = S1[rt << 1] + S1[rt << 1 | 1];        S2[rt] = S2[rt << 1] + S2[rt << 1 | 1];    }}void update(int L, int R, int d, int l, int r, int rt) {    if(L <= l && r <= R) {        cnt[rt] += d;        push_up(l, r, rt);        return;    }    int m = (l + r) >> 1;    if(L <= m) update(L, R, d, lson);    if(R > m) update(L, R, d, rson);    push_up(l, r, rt);}int main() {    int n, T, ansk = 0;    //freopen("input.txt", "r", stdin);    scanf("%d", &T);    while(T--) {        rear = 0;        memset(cnt, 0, sizeof(cnt));        memset(S1, 0, sizeof(S1));        memset(S2, 0, sizeof(S2));        scanf("%d", &n);        for(int i = 1; i <= n; i++) {            int x1, y1, x2, y2;            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);            A[++rear] = x1, A[++rear] = x2;            Q[i] = Que(y1, x1, x2, 1);            Q[i + n] = Que(y2, x1, x2, -1);        }        sort(Q + 1, Q + 1 + 2 * n);        sort(A + 1, A + 1 + rear);        rear = unique(A + 1, A + 1 + rear) - A - 1;        LL ans = 0;        int last = 0;        for(int i = 1; i <= 2 * n; i++) {            ans += (LL)(Q[i].top - last) * S2[1];            update(BS(Q[i].L), BS(Q[i].R) - 1, Q[i].sign, root);            last = Q[i].top;        }        printf("Case %d: %I64d\n", ++ansk, ans);    }    return 0;}

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