线段树 hdu3642 Get The Treasury

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不得不说，这是一题非常经典的体积并。。然而还是debug了2个多小时...

首先思路：按z的大小排序。

然后相当于扫描面一样，，从体积的最下方向上方扫描，遇到这个面

就将对应的两条线加入到set中，或者从set中删除，然后再对set中的所有边，求一次面积并

由于最后求出来的是至少有3个体积叠加的部分的体积。

所以需要维护3个节点，然后push_up会稍微啰嗦一点...

刚开始在set插入线段和删除线段的时候，搞错了线段所在面的编号，，然后一直样例都过不去也是醉了...

总的来说代码比较冗长需要足够细心才能敲对..

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>

using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define root 1,rear-1,1

int const MX = 4e3 + 5;

struct Side {
    int z, x1, y1, x2, y2, d, id;
    bool operator<(const Side &b)const {
        if(z == b.z) return d < b.d;
        return z < b.z;
    }
    Side() {}
    Side(int _z, int _id, int _x1, int _y1, int _x2, int _y2, int _d) {
        z = _z; x1 = _x1; y1 = _y1; x2 = _x2; y2 = _y2; d = _d; id = _id;
    }
} A[MX];

struct Line {
    int y, x1, x2, d, id;
    bool operator<(const Line &b)const {
        if(y == b.y) {
            if(d == b.d) return id < b.id;
            return d < b.d;
        }
        return y < b.y;
    }
    Line() {}
    Line(int _y, int _id, int _d, int _x1, int _x2) {
        y = _y; x1 = _x1; x2 = _x2; d = _d; id = _id;
    }
};

int X[MX], rear;
int S1[MX << 2], S2[MX << 2], S3[MX << 2], cnt[MX << 2];
set<Line>work;

int BS(int x) {
    int L = 1, R = rear, m;
    while(L <= R) {
        m = (L + R) >> 1;
        if(X[m] == x) return m;
        if(X[m] > x) R = m - 1;
        else L = m + 1;
    }
    return -1;
}

void push_up(int l, int r, int rt) {
    if(cnt[rt]) {
        S1[rt] = X[r + 1] - X[l];
        if(cnt[rt] == 1) {
            S2[rt] = S1[rt << 1] + S1[rt << 1 | 1];
            S3[rt] = S2[rt << 1] + S2[rt << 1 | 1];
        } else {
            S2[rt] = X[r + 1] - X[l];
            if(cnt[rt] == 2) {
                S3[rt] = S1[rt << 1] + S1[rt << 1 | 1];
            } else {
                S3[rt] = X[r + 1] - X[l];
            }
        }
    } else if(l == r) S1[rt] = S2[rt] = S3[rt] = 0;
    else {
        S1[rt] = S1[rt << 1] + S1[rt << 1 | 1];
        S2[rt] = S2[rt << 1] + S2[rt << 1 | 1];
        S3[rt] = S3[rt << 1] + S3[rt << 1 | 1];
    }
}

void update(int L, int R, int d, int l, int r, int rt) {
    if(L <= l && r <= R) {
        cnt[rt] += d;
        push_up(l, r, rt);
        return;
    }

    int m = (l + r) >> 1;
    if(L <= m) update(L, R, d, lson);
    if(R > m) update(L, R, d, rson);
    push_up(l, r, rt);
}

LL solve() {
    memset(S1, 0, sizeof(S1));
    memset(S2, 0, sizeof(S2));
    memset(cnt, 0, sizeof(cnt));

    LL ans = 0;
    int last = 0;
    for(set<Line>::iterator it = work.begin(); it != work.end(); it++) {
        ans += (LL)(it->y - last) * S3[1];
        update(BS(it->x1), BS(it->x2) - 1, it->d, root);
        last = it->y;
    }
    return ans;
}

int main() {
    //freopen("input.txt", "r", stdin);
    int T, n, ansk = 0;
    scanf("%d", &T);
    while(T--) {
        rear = 0;
        work.clear();

        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            int x1, y1, z1, x2, y2, z2;
            scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);
            A[i] = Side(z1, i, x1, y1, x2, y2, 1);
            A[i + n] = Side(z2, i, x1, y1, x2, y2, -1);
            X[++rear] = x1; X[++rear] = x2;
        }
        sort(A + 1, A + 1 + 2 * n);
        sort(X + 1, X + 1 + rear);
        rear = unique(X + 1, X + 1 + rear) - X - 1;

        LL ans = 0;
        int last = 0;
        printf("Case %d: ", ++ansk);
        for(int i = 1; i <= 2 * n; i++) {
            ans += (LL)(A[i].z - last) * solve();

            if(A[i].d > 0) {
                work.insert(Line(A[i].y1, A[i].id, 1, A[i].x1, A[i].x2));
                work.insert(Line(A[i].y2, A[i].id, -1, A[i].x1, A[i].x2));
            } else {
                work.erase(Line(A[i].y1, A[i].id, 1, A[i].x1, A[i].x2));
                work.erase(Line(A[i].y2, A[i].id, -1, A[i].x1, A[i].x2));
            }
            last = A[i].z;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

线段树 hdu3642 Get The Treasury

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原文地址：http://blog.csdn.net/qwb492859377/article/details/47441975