题意：

有一块田，上面有n个矩阵，每个矩阵对应一个权值，矩阵相交的部分取权值大的，问最后能获得多少值。

思路：

1. 权值问题可以转换为体积问题，因为相同格子里面权值大的胜，可以把权值看成是立方体的高，于是问题转换为立方体体积的并。

2. 和 get the treasure 题目类似，因为price变化的范围很小，于是先对高进行离散化 + 扫描。

3. 剩下的就是求 面积 * 高 了，对于面积，采取以前的常规手段，对 x 进行离散化，然后求面积的并就OK了。

#include <iostream>
#include <algorithm>
using namespace std;#define lhs l, m, rt << 1
#define rhs m + 1, r, rt << 1 | 1const int maxn = 61000;
int sum[maxn << 2], cnt[maxn << 2];
int price[10];
int xcord[maxn];struct Coord {int x1, y1, x2, y2, p;
} cord[30010] ;struct Segment {int l, r, h, v;Segment () { }Segment(int _l, int _r, int _h, int _v) : l(_l), r(_r), h(_h), v(_v) { }bool operator < (const Segment& other){if (h == other.h)return v > other.v;elsereturn h < other.h;}
} seg[maxn] ;void pushUp(int l, int r, int rt)
{if (cnt[rt])sum[rt] = xcord[r+1] - xcord[l];else if (l == r)sum[rt] = 0;elsesum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}void update(int beg, int end, int value, int l, int r, int rt)
{if (beg <= l && r <= end){cnt[rt] += value;pushUp(l, r, rt);return ;}int m = (l + r) >> 1;if (beg <= m)update(beg, end, value, lhs);if (end > m)update(beg, end, value, rhs);pushUp(l, r, rt);
}__int64 solve(int nn, int mm)
{__int64 ret = 0;sort(price + 1, price + mm + 1);for (int i = 1; i <= mm; ++i){int n = 0, m = 0;for (int j = 0; j < nn; ++j){if (cord[j].p >= price[i]){seg[n++] = Segment(cord[j].x1, cord[j].x2, cord[j].y1, 1);seg[n++] = Segment(cord[j].x1, cord[j].x2, cord[j].y2, -1);xcord[m++] = cord[j].x1;xcord[m++] = cord[j].x2;}}sort(seg, seg + n);sort(xcord, xcord + m);m = unique(xcord, xcord + m) - xcord;memset(sum, 0, sizeof(sum));memset(cnt, 0, sizeof(cnt));__int64 area = 0;for (int j = 0; j < n - 1; ++j){int beg = lower_bound(xcord, xcord + m, seg[j].l) - xcord;int end = lower_bound(xcord, xcord + m, seg[j].r) - xcord - 1;if (beg <= end)update(beg, end, seg[j].v, 0, m - 1, 1);area += (__int64)sum[1] * (seg[j+1].h - seg[j].h);}ret += area * (price[i] - price[i-1]);}return ret;
}int main()
{int cases, cc = 0;scanf("%d", &cases);while (cases--){int n, m;scanf("%d %d", &n, &m);price[0] = 0;for (int i = 1; i <= m; ++i)scanf("%d", &price[i]);for (int i = 0; i < n; ++i){scanf("%d %d %d %d %d", &cord[i].x1, &cord[i].y1, &cord[i].x2, &cord[i].y2, &cord[i].p);cord[i].p = price[cord[i].p];}printf("Case %d: %I64d\n", ++cc, solve(n, m));}return 0;
}