标签：void res may 矩形 single rate number long logs
题目链接：https://vjudge.net/problem/HDU-3642
InputThe first line of the input file contains a single integer t, the number of test cases, followed by the input data for each test case.
Each test case is given in some lines. In the first line there is an integer n (1 ≤ n ≤ 1000), the number of spots on the surface of the earth that he had detected. Then n lines follow, every line contains six integers x 1, y 1, z 1, x 2, y 2 and z2, separated by a space. The absolute value of x and y coordinates of the vertices is no more than 10 6, and that of z coordinate is no more than 500.
OutputFor each test case, you should output “Case a: b” in a single line. a is the case number, and b is the minimum volume of treasury. The case number is counted from one.
Sample Input
2 1 0 0 0 5 6 4 3 0 0 0 5 5 5 3 3 3 9 10 11 3 3 3 13 20 45
Sample Output
Case 1: 0 Case 2: 8
代码如下：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <vector> 7 #include <queue> 8 #include <stack> 9 #include <map> 10 #include <string> 11 #include <set> 12 using namespace std; 13 typedef long long LL; 14 const double EPS = 1e-8; 15 const int INF = 2e9; 16 const LL LNF = 2e18; 17 const int MAXN = 1e3+10; 18 19 int times[MAXN<<3]; 20 int one[MAXN<<3], two[MAXN<<3], more[MAXN<<3]; 21 int Z[MAXN<<1], X[MAXN<<1]; 22 23 struct Cube 24 { 25 int x1, y1, z1,x2, y2, z2; 26 }cube[MAXN]; 27 28 struct Line 29 { 30 int le, ri, h, id; 31 bool operator<(const Line &a)const{ 32 return h<a.h; 33 } 34 35 }line[MAXN<<1]; 36 37 void push_up(int u, int l, int r) 38 { 39 if(times[u]>=3) 40 { 41 more[u] = X[r] - X[l]; 42 two[u] = X[r] - X[l]; 43 one[u] = X[r] - X[l]; 44 } 45 else if(times[u]==2) 46 { 47 more[u] = (l+1==r)?0:(one[u*2]+one[u*2+1]); 48 two[u] = X[r] - X[l]; 49 one[u] = X[r] - X[l]; 50 } 51 else if(times[u]==1) 52 { 53 more[u] = (l+1==r)?0:(two[u*2]+two[u*2+1]); 54 two[u] = (l+1==r)?0:(one[u*2]+one[u*2+1]); 55 one[u] = X[r] - X[l]; 56 } 57 else 58 { 59 more[u] = (l+1==r)?0:(more[u*2]+more[u*2+1]); 60 two[u] = (l+1==r)?0:(two[u*2]+two[u*2+1]); 61 one[u] = (l+1==r)?0:(one[u*2]+one[u*2+1]); 62 } 63 } 64 65 void add(int u, int l, int r, int x, int y, int v) 66 { 67 if(x<=l && r<=y) 68 { 69 times[u] += v; 70 push_up(u, l, r); 71 return; 72 } 73 74 int mid = (l+r)>>1; 75 if(x<=mid-1) add(u*2, l, mid, x, y, v); 76 if(y>=mid+1) add(u*2+1, mid, r, x, y, v); 77 push_up(u, l, r); 78 } 79 80 int main() 81 { 82 int T, n; 83 scanf("%d", &T); 84 for(int kase = 1; kase<=T; kase++) 85 { 86 scanf("%d", &n); 87 int numZ = 0; 88 for(int i = 1; i<=n; i++) 89 { 90 scanf("%d%d%d", &cube[i].x1,&cube[i].y1,&cube[i].z1); 91 scanf("%d%d%d", &cube[i].x2,&cube[i].y2,&cube[i].z2); 92 Z[++numZ] = cube[i].z1; Z[++numZ] = cube[i].z2; 93 } 94 95 sort(Z+1, Z+1+numZ); 96 numZ = unique(Z+1, Z+1+numZ) - (Z+1); 97 98 LL volume = 0; 99 for(int i = 1; i<numZ; i++) 100 { 101 int numLine = 0, numX = 0; 102 for(int j = 1; j<=n; j++) 103 if(cube[j].z1<=Z[i] && Z[i+1]<=cube[j].z2) 104 { 105 line[++numLine].le = cube[j].x1; line[numLine].ri = cube[j].x2; 106 line[numLine].h = cube[j].y1; line[numLine].id = 1; 107 line[++numLine].le = cube[j].x1; line[numLine].ri = cube[j].x2; 108 line[numLine].h = cube[j].y2; line[numLine].id = -1; 109 X[++numX] = cube[j].x1; X[++numX] = cube[j].x2; 110 } 111 112 sort(line+1, line+1+numLine); 113 sort(X+1, X+1+numX); 114 numX = unique(X+1, X+1+numX) - (X+1); 115 116 memset(times, 0, sizeof(times)); 117 memset(more, 0, sizeof(more)); 118 memset(two, 0, sizeof(two)); 119 memset(one, 0, sizeof(one)); 120 121 LL area = 0; 122 for(int j = 1; j<numLine; j++) 123 { 124 int l = upper_bound(X+1, X+1+numX, line[j].le) - (X+1); 125 int r = upper_bound(X+1, X+1+numX, line[j].ri) - (X+1); 126 add(1, 1, numX, l, r, line[j].id); 127 area += 1LL*more[1]*(line[j+1].h-line[j].h); 128 } 129 volume += 1LL*area*(Z[i+1]-Z[i]); 130 } 131 printf("Case %d: %lld\n", kase, volume); 132 } 133 }
HDU3642 Get The Treasury —— 求矩形交体积 线段树 + 扫描线 + 离散化
标签：void res may 矩形 single rate number long logs
原文地址：http://www.cnblogs.com/DOLFAMINGO/p/7749513.html
