题意：给n个矩形，求矩形周长并
解法：跟求矩形面积并差不多，不过线段树节点记录的为：
len: 此区间线段长度
cover: 此区间是否被整个覆盖
lmark,rmark: 此区间左右端点是否被覆盖
num: 此区间分离开的线段的条数
重点在转移的地方，不难理解。
代码：
#include <iostream> #include <cmath> #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> using namespace std; #define N 10007 struct node { int cov,len,num,lmark,rmark; }tree[8*N]; struct Line { int y1,y2,x; int cov; }line[2*N]; int yy[2*N]; int cmp(Line ka,Line kb) { return ka.x < kb.x; } int bsearch(int l,int r,int x) { while(l <= r) { int mid = (l+r)/2; if(x == yy[mid]) return mid; if(x > yy[mid]) l = mid+1; else r = mid-1; } return l; } void build(int l,int r,int rt) { tree[rt].cov = tree[rt].len = tree[rt].num = tree[rt].lmark = tree[rt].rmark = 0; if(l == r-1) return; int mid = (l+r)/2; build(l,mid,2*rt); build(mid,r,2*rt+1); } void pushup(int l,int r,int rt) { if(tree[rt].cov) { tree[rt].len = yy[r]-yy[l]; tree[rt].lmark = tree[rt].rmark = 1; tree[rt].num = 1; } else if(l+1 == r) tree[rt].len = tree[rt].num = tree[rt].lmark = tree[rt].rmark = 0; else { tree[rt].len = tree[2*rt].len + tree[2*rt+1].len; tree[rt].lmark = tree[2*rt].lmark; tree[rt].rmark = tree[2*rt+1].rmark; tree[rt].num = tree[2*rt].num+tree[2*rt+1].num-tree[2*rt].rmark*tree[2*rt+1].lmark; //如果左右区间连续，那么可以合并，减少一个区间 } } void update(int l,int r,int aa,int bb,int cover,int rt) { if(aa <= l && bb >= r) { tree[rt].cov += cover; pushup(l,r,rt); return; } if(l+1 == r) return; int mid = (l+r)/2; if(aa <= mid) update(l,mid,aa,bb,cover,2*rt); if(bb > mid) update(mid,r,aa,bb,cover,2*rt+1); pushup(l,r,rt); } int main() { int n,m,cs = 1,i,j; int x1,y1,x2,y2; while(scanf("%d",&n)!=EOF) { m = 1; for(i=0;i<n;i++) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); line[m].x = x1,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = 1,yy[m++] = y1; line[m].x = x2,line[m].y1 = y1,line[m].y2 = y2,line[m].cov = -1,yy[m++] = y2; } m--; sort(yy+1,yy+m+1); int cnt = 2; for(i=2;i<=m;i++) { if(yy[i] != yy[i-1]) yy[cnt++] = yy[i]; } cnt--; build(1,cnt,1); sort(line+1,line+m+1,cmp); int ans = 0; int last = 0; for(i=1;i<=m;i++) { int L = bsearch(1,cnt,line[i].y1); int R = bsearch(1,cnt,line[i].y2); update(1,cnt,L,R,line[i].cov,1); ans += abs(last - tree[1].len); if(i == m) break; ans += tree[1].num*2*(line[i+1].x-line[i].x); last = tree[1].len; } printf("%d ",ans); } return 0; }