HDU 3642 Get The Treasury

体积交。给出若干长方体，求相交超过2次（3次及以上）的部分的总体积。对x，z分别离散化，注意pushup的处理，其他都是老生常谈了（最近队友总是爱用这个词）。

用sum[3][M]来表示的前区间的覆盖情况，sum[0]指覆盖一次，sum[1]指覆盖两次，sum[2]指覆盖多次。cnt[M]来记录当前区间的覆盖次数。

如果当前区间被覆盖了三次及以上，那么sum[0], sum[1]都为0，sum[2]为区间长度。

如果的前区间被覆盖了两次，那么sum[2]可以由左右孩子中的sum[0], sum[1]，sum[2]相加得到，sum[0]肯定是0了（至少两次），sum[1]用区间长度减去sum[2]即可。

如果的前区间被覆盖了一次，那么sum[2]由左右孩子中的sum[1]，sum[2]相加得到，sum[1]由左右孩子中的sum[0]相加得到，sum[0]是用区间长度减去另外两个。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<string>#include<queue>#include<map>///LOOP#define REP(i, n) for(int i = 0; i < n; i++)#define FF(i, a, b) for(int i = a; i < b; i++)#define FFF(i, a, b) for(int i = a; i <= b; i++)#define FD(i, a, b) for(int i = a - 1; i >= b; i--)#define FDD(i, a, b) for(int i = a; i >= b; i--)///INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)#define RFI(n) scanf("%lf", &n)#define RFII(n, m) scanf("%lf%lf", &n, &m)#define RFIII(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)#define RFIV(n, m, k, p) scanf("%lf%lf%lf%lf", &n, &m, &k, &p)#define RS(s) scanf("%s", s)///OUTPUT#define PN printf("\n")#define PI(n) printf("%d\n", n)#define PIS(n) printf("%d ", n)#define PS(s) printf("%s\n", s)#define PSS(s) printf("%s ", s)#define PC(n) printf("Case %d: ", n)///OTHER#define PB(x) push_back(x)#define CLR(a, b) memset(a, b, sizeof(a))#define CPY(a, b) memcpy(a, b, sizeof(b))#define display(A, n, m) {REP(i, n){REP(j, m)PIS(A[i][j]);PN;}}#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1using namespace std;typedef long long LL;typedef pair<int, int> P;const int MOD = 1e9+7;const int INFI = 1e9 * 2;const LL LINFI = 1e17;const double eps = 1e-6;const int N = 2222;const int M = N << 2;const int move[8][2] = {0, 1, 0, -1, 1, 0, -1, 0, 1, 1, 1, -1, -1, 1, -1, -1};struct node{    int x1, x2, y, k;    node(){};    node(int a, int b, int c, int d){x1 = a, x2 = b, y = c, k = d;};    bool operator < (const node p) const    {        return y < p.y;    }}s;struct point{    int x1, x2, y1, y2, z1, z2;}p[M];int cnt[M], Z[M], L, R;LL X[M], sum[3][M];vector<node> v[N];void pushup(int l, int r, int rt){    if(cnt[rt] > 2)    {        sum[2][rt] = X[r + 1] - X[l];        sum[0][rt] = sum[1][rt] = 0;    }    else if(cnt[rt] == 2)    {        sum[2][rt] = 0;        REP(i, 3)sum[2][rt] += sum[i][rt << 1] + sum[i][rt << 1 | 1];        sum[1][rt] = X[r + 1] - X[l] - sum[2][rt];        sum[0][rt] = 0;    }    else if(cnt[rt] == 1)    {        sum[2][rt] = sum[2][rt << 1] + sum[2][rt << 1 | 1] + sum[1][rt << 1] + sum[1][rt << 1 | 1];        sum[1][rt] = sum[0][rt << 1] + sum[0][rt << 1 | 1];        sum[0][rt] = X[r + 1] - X[l] - sum[2][rt] - sum[1][rt];    }    else if(l == r)REP(i, 3)sum[i][rt] = 0;    else REP(i, 3)sum[i][rt] = sum[i][rt << 1] + sum[i][rt << 1 | 1];}void update(int c, int l, int r, int rt){    if(L <= l && r <= R)    {        cnt[rt] += c;        pushup(l, r, rt);        return;    }    int m = (l + r) >> 1;    if(L <= m)update(c, lson);    if(R > m)update(c, rson);    pushup(l, r, rt);}int main(){    //freopen("input.txt", "r", stdin);    //freopen("output.txt", "w", stdout);    int t, n, mx, mz, k;    LL ans, tmp;    RI(t);    FFF(cas, 1, t)    {        RI(n);        REP(i, N)v[i].clear();        mx = mz = ans = tmp = 0;        REP(i, n)        {            RIII(p[i].x1, p[i].y1, p[i].z1);            RIII(p[i].x2, p[i].y2, p[i].z2);            X[mx++] = p[i].x1;            X[mx++] = p[i].x2;            Z[mz++] = p[i].z1;            Z[mz++] = p[i].z2;        }        sort(X, X + mx);        mx = unique(X, X + mx) - X;        sort(Z, Z + mz);        mz = unique(Z, Z + mz) - Z;        REP(i, n)        {            L = lower_bound(Z, Z + mz, p[i].z1) - Z;            R = lower_bound(Z, Z + mz, p[i].z2) - Z;            FF(j, L, R)            {                v[j].PB(node(p[i].x1, p[i].x2, p[i].y1, 1));                v[j].PB(node(p[i].x1, p[i].x2, p[i].y2, -1));            }        }        REP(i, mz)if(v[i].size())sort(v[i].begin(), v[i].end());        PC(cas);        CLR(cnt, 0);CLR(sum, 0);        REP(i, mz)        {            k = v[i].size();            REP(j, k)            {                L = lower_bound(X, X + mx, v[i][j].x1) - X;                R = lower_bound(X, X + mx, v[i][j].x2) - X - 1;                update(v[i][j].k, 0, mx - 1, 1);                if(j != k - 1)tmp += sum[2][1] * LL(v[i][j + 1].y - v[i][j].y);            }            if(i != mz - 1)ans += tmp * LL(Z[i + 1] - Z[i]);            tmp = 0;        }        printf("%I64d\n", ans);    }    return 0;}