hdu 4533(树状数组)

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题意：给出n个矩形的左下角和右上角坐标，问矩形(0,0),(ti,ti)内矩形面积是多少，重叠也算。
题解：参照http://blog.csdn.net/shiqi_614/article/details/8869676这里的题解，把所有矩形关于t的重叠面积计算式中的系数用树状数组维护，很机智。

#include <stdio.h>#include <string.h>#include <iostream>#define ll long longusing namespace std;const int N = 200005;int n, q;ll S[3][N], x1, y1, x2, y2;int lowbit(int x) { return x & (-x); }void modify(int st, int en, ll v, int flag) {    for (int i = st; i < N; i += lowbit(i))        S[flag][i] += v;    for (int i = en + 1; i < N; i += lowbit(i))        S[flag][i] -= v;}ll query(int x, int flag) {    ll ret = 0;    for (int i = x; i > 0; i -= lowbit(i))        ret += S[flag][i];    return ret;}int main() {    int t;    scanf("%d", &t);    while (t--) {        memset(S, 0, sizeof(S));        scanf("%d", &n);        for (int i = 0; i < n; i++) {            scanf("%lld%lld%lld%lld", &x1, &y1, &x2, &y2);            ll temp1 = max(x1, y1), temp2 = min(x2, y2);            if (temp1 < temp2) {                modify(temp1, temp2, 1, 0);                modify(temp1, temp2, -x1 - y1, 1);                modify(temp1, temp2, x1 * y1, 2);            }            if (y2 > x2) {                temp1 = max(x2, y1) + 1;                modify(temp1, y2, x2 - x1, 1);                modify(temp1, y2, x1 * y1 - x2 * y1, 2);            }            if (x2 > y2) {                temp1 = max(y2, x1) + 1;                modify(temp1, x2, y2 - y1, 1);                modify(temp1, x2, x1 * y1 - x1 * y2, 2);            }            modify(max(x2, y2) + 1, N, (y2 - y1) * (x2 - x1), 2);        }        scanf("%d", &q);        while (q--) {            ll tt;            scanf("%lld", &tt);            printf("%lld\n", tt * tt * query(tt, 0) + tt * query(tt, 1) + query(tt, 2));        }    }    return 0;}

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