看的HH的题解。。周长有两部分组成，横着和竖着的，横着通过，sum[1] - last来计算，竖着的通过标记，记录有多少段。
1 #include <cstdio> 2 #include <cstring> 3 #include <string> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 #define maxn 10000 8 #define lson l , m, rt<<1 9 #define rson m+1, r,rt<<1|1 10 int que[maxn*4]; 11 int sum[maxn*4]; 12 int cnt[maxn*4]; 13 bool lbd[maxn*4] , rbd[maxn*4]; 14 int numseg[maxn*4]; 15 struct node 16 { 17 int lx,rx,y; 18 int s; 19 node() {} 20 node(int a,int b,int c,int d):lx(a),rx(b),y(c),s(d) {} 21 bool operator < (const node &S) const 22 { 23 if(y == S.y) return s > S.s; 24 return y < S.y; 25 } 26 } mat[maxn]; 27 int bin(int x,int n) 28 { 29 int str,end,mid; 30 str = 0,end = n; 31 while(str <= end) 32 { 33 mid = (str+end)/2; 34 if(que[mid] == x) 35 return mid; 36 else if(que[mid] > x) 37 end = mid - 1; 38 else 39 str = mid + 1; 40 } 41 return mid; 42 } 43 void pushup(int rt,int l,int r) 44 { 45 if(cnt[rt]) 46 { 47 lbd[rt] = rbd[rt] = 1; 48 sum[rt] = que[r+1] - que[l]; 49 numseg[rt] = 2; 50 } 51 else if (l == r) 52 { 53 sum[rt] = numseg[rt] = lbd[rt] = rbd[rt] = 0; 54 } 55 else 56 { 57 lbd[rt] = lbd[rt<<1]; 58 rbd[rt] = rbd[rt<<1|1]; 59 sum[rt] = sum[rt<<1] + sum[rt<<1|1]; 60 numseg[rt] = numseg[rt<<1] + numseg[rt<<1|1]; 61 if (lbd[rt<<1|1] && rbd[rt<<1]) numseg[rt] -= 2; 62 } 63 } 64 void update(int L,int R,int c,int l,int r,int rt) 65 { 66 if(L <= l&&r <= R) 67 { 68 cnt[rt] += c; 69 pushup(rt,l,r); 70 return ; 71 } 72 int m = (l+r)>>1; 73 if(L <= m) 74 update(L,R,c,lson); 75 if(R > m) 76 update(L,R,c,rson); 77 pushup(rt,l,r); 78 } 79 int main() 80 { 81 int n,num,i,l,r; 82 int a,b,c,d; 83 while(scanf("%d",&n)!=EOF) 84 { 85 if(!n) break; 86 num = 0; 87 for(i = 0; i < n; i ++) 88 { 89 scanf("%d%d%d%d",&a,&b,&c,&d); 90 que[num] = a; 91 mat[num ++] = node(a,c,b,1); 92 que[num] = c; 93 mat[num ++] = node(a,c,d,-1); 94 } 95 sort(que,que+num); 96 sort(mat,mat+num); 97 int k = 1; 98 for(i = 1; i < num; i ++) 99 { 100 if(que[i] != que[i-1]) 101 que[k++] = que[i]; 102 } 103 int ans = 0,last = 0; 104 memset(cnt,0,sizeof(cnt)); 105 memset(sum,0,sizeof(sum)); 106 memset(lbd,0,sizeof(lbd)); 107 memset(rbd,0,sizeof(rbd)); 108 for(i = 0; i < num; i ++) 109 { 110 l = bin(mat[i].lx,k-1); 111 r = bin(mat[i].rx,k-1)-1; 112 if(l <= r) 113 update(l,r,mat[i].s,0,k-1,1); 114 ans += numseg[1] * (mat[i+1].y - mat[i].y); 115 ans += abs(sum[1] - last); 116 last = sum[1]; 117 } 118 printf("%d ",ans); 119 } 120 return 0; 121 }