HDU 3255 Farming

体积并。在某些矩形内撒种子，重复的地方取价值最高的那个，每个点只能存活一个，求总价值。最先想到的是二维线段树，区间修改，然后统计各种植物的个数，再乘以价值，得到答案，感觉挺麻烦的。然后发现，把价值当高度，就变成体积并了。先求一层的面积，再乘以两层之间的高度差，如此累加。每层的初始面积都是0，上一层的不要了。对于一个长方体，能到哪层，就在那层留下它的足迹。用vector来记录每层的线段。注意会超int。

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#include<vector>#include<string>#include<queue>#include<cmath>#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 double pi = acos(-1.0);const int N = 33333;const int M = N << 3;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 price{    int v, pos;    bool operator < (const price x) const    {        return v < x.v;    }}p[5];int cnt[M], f[5], X[M], L, R;LL sum[M];vector<node> v[5];void pushup(int l, int r, int rt){    if(cnt[rt])sum[rt] = X[r + 1] - X[l];    else if(l == r)sum[rt] = 0;    else sum[rt] = sum[rt << 1] + sum[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);    int t, n, m, k, a, b, c, d, e;    LL ans, tmp;    RI(t);    FFF(cas, 1, t)    {        RII(n, m);        REP(i, m)        {            RI(p[i].v);            p[i].pos = i + 1;        }        sort(p, p + m);        REP(i, m)f[p[i].pos] = i + 1;        REP(i, 5)v[i].clear();        k = ans = 0;        REP(i, n)        {            RV(a, b, c, d, e);            X[k++] = a;            X[k++] = c;            s = node(a, c, b, 1);            FFF(j, 1, f[e])v[j].PB(s);            s = node(a, c, d, -1);            FFF(j, 1, f[e])v[j].PB(s);        }        FFF(i, 1, m)sort(v[i].begin(), v[i].end());        sort(X, X + k);        k = unique(X, X + k) - X;        PC(cas);        CLR(sum, 0);        CLR(cnt, 0);        FFF(i, 1, m)        {            e = v[i].size();            if(!e)continue;            tmp = 0;            REP(j, e)            {                L = lower_bound(X, X + k, v[i][j].x1) - X;                R = lower_bound(X, X + k, v[i][j].x2) - X - 1;                update(v[i][j].k, 0, k - 1, 1);                if(j < e)tmp += sum[1] * (v[i][j + 1].y - v[i][j].y);            }            if(i == 1)ans += tmp * LL(p[i - 1].v);            else ans += tmp * LL(p[i - 1].v - p[i - 2].v);        }        printf("%I64d\n", ans);    }    return 0;}