[Picture HDU - 1828 scanning lines] []

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  This question is seeking the value of the circumference of the synthesis block may overlap these small blocks may be composed of many, there would be a simple yet sophisticated approach is to run the two vertical and horizontal scanning lines, obtained final and two perimeter, but this approach seems a bit complicated, we can run only once tree line (scanning line) to do such a request.

  Ideas: the problem is that we have to go to know how many you can put up to the edge, there is between the two scan lines, independent of how many squares where? We seek an independent box number, multiplied by the number of edges is the final vertical 2, then the problem is solved vertical edge of the first we solve. Next is the length of the lateral side of Solving, we each scanning line, it represents the current process line to how long interval, then we find a difference between the on state, the current is not covered increase out of the side of the perimeter. Thus, we get the final total will add up to the perimeter of.

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cmath>
  4 #include <string>
  5 #include <cstring>
  6 #include <algorithm>
  7 #include <limits>
  8 #include <vector>
  9 #include <stack>
 10 #include <queue>
 11 #include <set>
 12 #include <map>
 13 #define lowbit(x) ( x&(-x) )
 14 #define pi 3.141592653589793
 15 #define e 2.718281828459045
 16 #define INF 0x3f3f3f3f
 17 #define HalF (l + r)>>1
 18 #define lsn rt<<1
 19 #define rsn rt<<1|1
 20 #define Lson lsn, l, mid
 21 #define Rson rsn, mid+1, r
 22 #define QL Lson, ql, qr
 23 #define QR Rson, ql, qr
 24 #define myself rt, l, r
 25 using namespace std;
 26 typedef unsigned long long ull;
 27 typedef long long ll;
 28 const int maxN = 1e4 + 7;
 29 int N, tot, X[maxN], _UP;
 30 struct node
 31 {
 32     int lx, rx, y, val;
 33     node(int a=0, int b=0, int c=0, int d=0):lx(a), rx(b), y(c), val(d) {}
 34 }line[maxN];
 35 bool cmp(node e1, node e2) { return e1.y < e2.y; }
 36 struct tree
 37 {
 38     int sum, siz, num;
 39     bool lc, rc;
 40     tree(int a=0, int b=0, int f=0, bool c=false, bool d=false):sum(a), siz(b), num(f), lc(c), rc(d) {}
 41     void clear() { sum = siz = num = 0; lc = rc = false; }
 42 }t[maxN<<2];
 43 inline void buildTree(int rt, int l, int r)
 44 {
 45     t[rt].clear();
 46     if(l == r) return;
 47     int mid = HalF;
 48     buildTree(Lson);
 49     buildTree(Rson);
 50 }
 51 inline void pushup(int rt, int l, int r)
 52 {
 53     if(t[rt].siz)
 54     {
 55         t[rt].sum = X[r + 1] - X[l];
 56         t[rt].lc = t[rt].rc = true;
 57         t[rt].num = 1;
 58     }
 59     else if(l == r)
 60     {
 61         t[rt].sum = t[rt].num = 0;
 62         t[rt].lc = t[rt].rc = false;
 63     }
 64     else
 65     {
 66         t[rt].sum = t[lsn].sum + t[rsn].sum;
 67         t[rt].lc = t[lsn].lc;   t[rt].rc = t[rsn].rc;
 68         t[rt].num = t[lsn].num + t[rsn].num - (t[lsn].rc && t[rsn].lc);
 69     }
 70 }
 71 inline void update(int rt, int l, int r, int ql, int qr, int val)
 72 {
 73     if(ql <= l && qr >= r)
 74     {
 75         t[rt].siz += val;
 76         pushup(myself);
 77         return;
 78     }
 79     int mid = HalF;
 80     if(qr <= mid) update(QL, val);
 81     else if(ql > mid) update(QR, val);
 82     else { update(QL, val); update(QR, val); }
 83     pushup(myself);
 84 }
 85 inline void init()
 86 {
 87     tot = 0;
 88 }
 89 int main()
 90 {
 91     while(scanf("%d", &N) != EOF)
 92     {
 93         init();
 94         int lx, ly, rx, ry;
 95         for(int i=1; i<=N; i++)
 96         {
 97             scanf("%d%d%d%d", &lx, &ly, &rx, &ry);
 98             line[++tot] = node(lx, rx, ly, 1);
 99             X[tot] = lx;
100             line[++tot] = node(lx, rx, ry, -1);
101             X[tot] = rx;
102         }
103         sort(X + 1, X + tot + 1);
104         sort(line + 1, line + tot + 1, cmp);
105         _UP = (int)(unique(X + 1, X + tot + 1) - X - 1);
106         buildTree(1, 1, _UP);
107         int ans = 0, las = 0;
108         for(int i=1, l, r; i<tot; i++)
109         {
110             l = (int)(lower_bound(X + 1, X + _UP + 1, line[i].lx) - X);
111             r = (int)(lower_bound(X + 1, X + _UP + 1, line[i].rx) - X - 1);
112             update(1, 1, _UP, l, r, line[i].val);
113             ans += abs(las - t[1].sum);
114             las = t[1].sum;
115             ans += (line[i + 1].y - line[i].y) * (t[1].num << 1);
116         }
117         ans += line[tot].rx - line[tot].lx;
118         printf("%d\n", ans);
119     }
120     return 0;
121 }
122 /*
123 7
124 -15 0 5 10
125 -5 8 20 25
126 15 -4 24 14
127 0 -6 16 4
128 2 15 10 22
129 30 10 36 20
130 34 0 40 16
131 ans = 228
132 */

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