hdu 1542 线段树应用 附几组测试数据

这题本来可以不用线段树做的，因为数据比较弱，直接离散化即可。但为了练习线段树，还是用线段树打了，可是却WA了一整天！调了四个小时以后仍没找出错误，让竹子帮我测试，测出了一处错误，又去北大讨论版找了些数据测，终于过了。

我的代码如下，第一次写这类题目，写得比较乱，以后再改进吧。

/*
 * hdu1542/linux.cpp
 * Created on: 2011-9-1
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef struct CNode {
    int L, R;
    int covers;
    double lastx;
    CNode *pLeft, *pRight;
} CNode;

typedef struct {
    double x;
    double yi;
    double ya;
    int cover;
} Line;

const int MAXN = 205;
const int MAX_NODE = 4 * MAXN;

CNode Tree[MAX_NODE];
Line lines[2 * MAXN];

double ys[2 * MAXN], ans;
int N, N2, nCount;

inline bool operator<(const Line &a, const Line &b) {
    return a.x < b.x;
}

//建立线段树
void BuildTree(CNode *pRoot, int L, int R) {
    pRoot->L = L;
    pRoot->R = R;
    pRoot->covers = 0;
    pRoot->lastx = 0;
    if (L == R) {
        return;
    }
    nCount++;
    pRoot->pLeft = Tree + nCount;
    nCount++;
    pRoot->pRight = Tree + nCount;
    int mid = (L + R) / 2;
    BuildTree(pRoot->pLeft, L, mid);
    BuildTree(pRoot->pRight, mid + 1, R);
}

//插入数据
void Update(CNode *pRoot, double i, double v, double x, int cover) {
    if (pRoot->L == pRoot->R) {
        if (pRoot->covers) {
            ans += (ys[pRoot->R] - ys[pRoot->L - 1]) * (x - pRoot->lastx);
        }
        pRoot->covers += cover;
        pRoot->lastx = x;
        return;
    }
    int mid = (pRoot->L + pRoot->R) / 2;
    if (pRoot->covers) {
        Update(pRoot->pLeft, ys[pRoot->L - 1], ys[mid], pRoot->lastx,
                pRoot->covers);
        Update(pRoot->pRight, ys[mid], ys[pRoot->R], pRoot->lastx,
                pRoot->covers);
        pRoot->covers = 0;
    }
    if (ys[mid] > i && ys[mid] < v) {
        Update(pRoot->pLeft, i, ys[mid], x, cover);
        Update(pRoot->pRight, ys[mid], v, x, cover);
    } else if (ys[mid] >= v) {
        Update(pRoot->pLeft, i, v, x, cover);
    } else {
        Update(pRoot->pRight, i, v, x, cover);
    }
}

void work() {
    int T = 0;
    double xi, yi, xa, ya;
    while (scanf("%d", &N) == 1 && N > 0) {
        N2 = 0;
        for (int i = 0; i < N; i++) {
            scanf("%lf%lf%lf%lf", &xi, &yi, &xa, &ya);
            lines[N2].yi = yi;
            lines[N2].ya = ya;
            lines[N2].cover = 1;
            lines[N2].x = xi;
            ys[N2++] = yi;
            lines[N2].yi = yi;
            lines[N2].ya = ya;
            lines[N2].cover = -1;
            lines[N2].x = xa;
            ys[N2++] = ya;
        }
        nCount = 0;
        BuildTree(Tree, 1, N2 - 1);
        sort(ys, ys + N2);
        sort(lines, lines + N2);
        ans = 0;
        for (int i = 0; i < N2; i++) {
            Update(Tree, lines[i].yi, lines[i].ya, lines[i].x, lines[i].cover);
        }
        printf("Test case #%d\nTotal explored area: %.2f\n\n", ++T, ans);
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    work();
    return 0;
}

　　

下面是我的测试数据，希望对以后做此题不过者有用。

输入数据：

2

1 2 6 3

3 1 4 5

5

137 10 275 20

296 201 306 275

265 105 497 115

233 74 402 84

296 10 402 147

5

1.37 0.10 2.75 0.20

2.96 2.01 3.06 2.75

2.65 1.05 4.97 1.15

2.33 0.74 4.02 0.84

2.96 0.10 4.02 1.47

3

1 1 6 7

4 2 11 5

2 0 8 3

3

1 1 3 3

2 2 6 6

0 0 4 4

3

3 3 4 4

1 1 5 5

2 2 3 3

3

1 1 4 4

2 2 3 3

10 10 20 20

2

1 1 4 4

2 2 3 3

3

15 15 25 25.5

10 10 20 20

15 15 20 20

2

1 1 2 2

2 2 3 3

1

1 1 2 2

2

10 10 20 20

10 20 20 30

3

10 10 20 20

15 15 25 25.5

30 10 40 20

2

15 15 25 25.5

10 10 20 20

8

9.33 19.52 30.44 24.39

8.06 29.38 9.75 51.77

22.38 15.06 32.35 33.30

30.66 19.20 44.13 19.62

17.93 15.38 39.35 41.58

9.65 6.47 29.80 10.39

26.20 24.61 40.63 35.85

25.25 12.51 29.80 36.17

8

157.98 278.30 246.25 501.22

152.89 15.06 355.43 214.74

61.53 56.12 141.85 96.97

21.43 303.77 250.07 329.65

5.83 115.65 306.10 211.88

275.44 132.52 353.21 276.49

149.71 175.49 271.40 379.94

127.42 32.57 187.05 282.22

0

输出数据如下

Test case #1

Total explored area: 8.00



Test case #2

Total explored area: 18532.00



Test case #3

Total explored area: 1.85



Test case #4

Total explored area: 53.00



Test case #5

Total explored area: 28.00



Test case #6

Total explored area: 16.00



Test case #7

Total explored area: 109.00



Test case #8

Total explored area: 9.00



Test case #9

Total explored area: 180.00



Test case #10

Total explored area: 2.00



Test case #11

Total explored area: 1.00



Test case #12

Total explored area: 200.00



Test case #13

Total explored area: 280.00



Test case #14

Total explored area: 180.00



Test case #15

Total explored area: 751.10



Test case #16

Total explored area: 99910.16