hdu 3642 体积并
题意：求三个矩形体积的并
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枚举z
1 #include<stdio.h> 2 #include<iostream> 3 #include<stdlib.h> 4 #include<string.h> 5 #include<algorithm> 6 #include<vector> 7 #include<math.h> 8 #include<map> 9 #pragma comment(linker, "/STACK:1024000000,1024000000") 10 using namespace std; 11 #define maxn 1100 12 #define mem(a,b) (memset(a),b,sizeof(a)) 13 #define lmin 1 14 #define rmax len 15 #define lson l,(l+r)/2,rt<<1 16 #define rson (l+r)/2+1,r,rt<<1|1 17 #define root lmin,rmax,1 18 #define now l,r,rt 19 #define int_now int l,int r,int rt 20 #define INF 99999999 21 #define LL long long 22 #define mod 10007 23 #define eps 1e-6 24 #define zero(x) (fabs(x)<eps?0:x) 25 #define LL __int64 26 map<int,int>mp; 27 int du[maxn*2]; 28 struct lines 29 { 30 int x; 31 int y,yy; 32 int z,zz; 33 int leap; 34 friend bool operator <(const lines &a,const lines &b) 35 { 36 return a.x<b.x; 37 } 38 } line[maxn*2]; 39 int num[maxn*4*2]; 40 int sum[maxn*4*2]; 41 int kum[maxn*4*2]; 42 int cover[maxn*4*2]; 43 void push_down(int_now) 44 { 45 46 } 47 void push_up(int_now) 48 { 49 int len=du[r+1]-du[l]; 50 if(cover[rt]==0) 51 { 52 num[rt]=num[rt<<1]+num[rt<<1|1]; 53 sum[rt]=sum[rt<<1]+sum[rt<<1|1]; 54 kum[rt]=kum[rt<<1]+kum[rt<<1|1]; 55 } 56 if(cover[rt]==1) 57 { 58 num[rt]=len; 59 sum[rt]=num[rt<<1]+num[rt<<1|1]; 60 kum[rt]=sum[rt<<1]+sum[rt<<1|1]; 61 } 62 if(cover[rt]==2) 63 { 64 num[rt]=len; 65 sum[rt]=len; 66 kum[rt]=num[rt<<1]+num[rt<<1|1]; 67 } 68 if(cover[rt]>=3) 69 { 70 num[rt]=len; 71 sum[rt]=len; 72 kum[rt]=len; 73 } 74 } 75 void creat() 76 { 77 memset(cover,0,sizeof(cover)); 78 memset(num,0,sizeof(num)); 79 memset(sum,0,sizeof(sum)); 80 memset(kum,0,sizeof(kum)); 81 } 82 void updata(int ll,int rr,int x,int_now) 83 { 84 if(ll>r||rr<l)return; 85 if(ll<=l&&rr>=r) 86 { 87 cover[rt]+=x; 88 push_up(now); 89 return; 90 } 91 updata(ll,rr,x,lson); 92 updata(ll,rr,x,rson); 93 push_up(now); 94 } 95 int main() 96 { 97 int T,cas; 98 scanf("%d",&T); 99 cas=0; 100 while(T--) 101 { 102 cas++; 103 int n,x,y,z,xx,yy,zz; 104 mp.clear(); 105 scanf("%d",&n); 106 LL ls=1; 107 du[0]=-1000010; 108 for(int i=1; i<=n; i++) 109 { 110 scanf("%d%d%d%d%d%d",&x,&y,&z,&xx,&yy,&zz); 111 line[i*2-1].x=x; 112 line[i*2-1].y=y; 113 line[i*2-1].yy=yy; 114 line[i*2-1].z=z; 115 line[i*2-1].zz=zz; 116 line[i*2-1].leap=1; 117 line[i*2].x=xx; 118 line[i*2].y=y; 119 line[i*2].yy=yy; 120 line[i*2].z=z; 121 line[i*2].zz=zz; 122 line[i*2].leap=-1; 123 du[ls++]=y; 124 du[ls++]=yy; 125 } 126 sort(line+1,line+n*2+1); 127 sort(du+1,du+ls); 128 int len=1; 129 for(int i=1; i<ls; i++) 130 { 131 if(du[i]!=du[i-1]) 132 { 133 mp[du[i]]=len; 134 du[len++]=du[i]; 135 } 136 } 137 len-=2; 138 LL are=0; 139 int st=0; 140 creat(); 141 for(int j=-501; j<501; j++) 142 { 143 st=0; 144 for(int i=1; i<=n*2; i++) 145 { 146 int l,r; 147 l=mp[line[i].y]; 148 r=mp[line[i].yy]; 149 if(line[i].z>j||line[i].zz<=j)continue; 150 LL x,y; 151 x=(LL)kum[1]; 152 y=(LL)line[i].x-st; 153 are+=x*y; 154 updata(l,r-1,line[i].leap,root); 155 st=line[i].x; 156 } 157 } 158 printf("Case %d: %I64d\n",cas,are); 159 } 160 return 0; 161 }
