hdu 4533 威威猫系列故事——晒被子

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看了网上各大神的解题，才发现不进行数学推导怎么都会超时。。。

选了个最简单粗暴的算法，对每个四边形的四个点(x,y)，若max(x, y) <= t，则加上(t,t)与(x,y)组成的矩形，否则减去，最后通过合并系数来简化求值操作。。。

#include<stdio.h>#include<algorithm>using namespace std;typedef __int64 ll;const int MAX = 50005;struct node{    ll x, y;    int maxnum;    void getmx()    {        maxnum = x>y?x:y;    }}data1[MAX], data2[MAX];int num1, num2;bool cmp(node &a, node &b){    return a.maxnum < b.maxnum;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);#endif    int t, n, x;    int i;    ll j;    scanf("%d", &t);    while (t--)    {        scanf("%d", &n);        num1 = num2 = 0;        node tpNode1, tpNode2;        for (i = 0; i< n; i++)        {            scanf("%I64d%I64d%I64d%I64d", &tpNode1.x, &tpNode1.y, &tpNode2.x, &tpNode2.y);            tpNode1.getmx(); tpNode2.getmx();            data1[num1++] = tpNode1;            data1[num1++] = tpNode2;            j = tpNode1.y;            tpNode1.y = tpNode2.y;            tpNode2.y = j;            tpNode1.getmx(); tpNode2.getmx();            data2[num2++] = tpNode1;            data2[num2++] = tpNode2;        }        sort(data1, data1+num1, cmp);        sort(data2, data2+num2, cmp);        int bg1 = 0, bg2 = 0;        ll now;        ll res = 0;        ll aa = 0, bb = 0, tt = 0;        scanf("%d", &x);        while (x--)        {            scanf("%I64d", &now);            for (; data1[bg1].maxnum <= now && bg1 <= num1; bg1++)            {                aa -= data1[bg1].x+data1[bg1].y;                bb += data1[bg1].x*data1[bg1].y;                tt++;            }            for ( ; data2[bg2].maxnum <= now && bg2 <= num2; bg2++)            {                aa += data2[bg2].x+data2[bg2].y;                bb -= data2[bg2].x*data2[bg2].y;                tt--;            }            res = tt*now*now+aa*now+bb;            printf("%I64d\n", res);        }    }    return 0;}