FZU 2187 回家种地（矩形面积并）

来源：互联网发布：python文本分析和提取编辑：程序博客网时间：2024/05/22 07:52

题意：

求的是只覆盖了一次的面积并是多少。

解析：

相对于普通的面积并来说，就是求cover==1的长度，去掉了cover>=1的长度。
参考了网络上的做法，用sum维护当前区间cover>=1的长度，single维护cover==1的长度。
其他地方和普通的面积并一模一样，只是在pushUp的时候稍微多了对single的维护而已

my code

#include <cstdio>#include <cstring>#include <algorithm>#define ls (o<<1)#define rs (o<<1|1)#define lson ls, L, M#define rson rs, M+1, Rusing namespace std;typedef __int64 ll;const int maxn = (int)2e5 + 5;struct Line {    ll l, r, h, d;    Line() {}    Line(ll l, ll r, ll h, ll d) : l(l), r(r), h(h), d(d) {}} line[maxn];bool cmp(Line a, Line b) {    return a.h < b.h;}int n, m, tot;ll X[maxn];ll sum[maxn<<2], single[maxn<<2];int cover[maxn<<2];void build(int o, int L, int R) {    sum[o] = single[o] = cover[o] = 0;    if(L == R) return ;    int M = (L + R)/2;    build(lson);    build(rson);}inline void pushUp(int o, int L, int R) {    if(cover[o]) {        sum[o] = X[R+1] - X[L];        if(cover[o] == 1) single[o] = sum[o] - sum[ls] - sum[rs];        else single[o] = 0;    }else if(L == R) sum[o] = single[o] = 0;    else {        sum[o] = sum[ls] + sum[rs];        single[o] = single[ls] + single[rs];    }}void modify(int o, int L, int R, int ql, int qr, int d) {    if(ql <= L && R <= qr) {        cover[o] += d;        pushUp(o, L, R);        return ;    }    int M = (L + R)/2;    if(ql <= M) modify(lson, ql, qr, d);    if(qr > M) modify(rson, ql, qr, d);    pushUp(o, L, R);}int find(int x) {      int L = 0, R = tot-1;     while(L <= R) {        int M = (L + R) >> 1;          if(X[M] == x) return M;        if(X[M] < x) L = M + 1;        else R = M - 1;    }    return -1;  }int main() {    int T, cas = 1;    int x1, y1, x2, y2;    scanf("%d", &T);    while(T--) {        m = tot = 0;        scanf("%d", &n);            for(int i = 0; i < n; i++) {            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);            line[m] = Line(x1, x2, y1, 1);            X[m++] = x1;            line[m] = Line(x1, x2, y2, -1);            X[m++] = x2;        }        sort(X, X+m);        tot = unique(X, X+m) - X;        sort(line, line+m, cmp);        build(1, 0, tot-1);        ll ans = 0;        int ql, qr;        for(int i = 0; i < m; i++) {            ql = find(line[i].l);            qr = find(line[i].r) - 1;            modify(1, 0, tot-1, ql, qr, line[i].d);            ans += single[1] * (line[i+1].h - line[i].h);        }        printf("Case %d: %I64d\n", cas++, ans);    }    return 0;}

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