http://poj.org/problem?id=1177
给若干矩形，求被覆盖的区域的周长。

将 y 坐标离散化后，按 x 坐标进行扫描。用线段树维护两个东西，当前竖线的叠加长度 len 和 条数 cnt 。 前一个用来计算竖直方向的周长部分，后一个用来计算水平方向的。
用 left[node] 和 right[node] 来记录每个结点左端和右端是否被覆盖，用来维护 cnt

重叠的边也不能计算，这反应在对扫描线的排序上，两线重叠时入边应在出边之前。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;#define CHD int lc = node<<1,rc = node<<1|1;
#define MID int mid = (L+R)>>1;
#define debug(x) cout<<"debug "<<x<<endl;
#define rep(i,f,t) for(int i = (f),_end =(t); i <= _end; ++i)struct Node{int x;int y1,y2;int flag; //入边为1，出边为-1Node(int x,int y1,int y2,int f):x(x),y1(y1),y2(y2),flag(f){}bool operator< (const Node &n2)const{if(x == n2.x)return flag > n2.flag;//入边在出边前return x < n2.x;}
};
vector<Node> line;
vector<int> vs;
const int maxn = 10005;struct sgt{int len[maxn<<2];int cnt[maxn<<2];int cov[maxn<<2];int left[maxn<<2];int right[maxn<<2];void maintain(int node,int L,int R){if(cov[node] > 0){len[node] = vs[R]-vs[L-1];cnt[node] = 1;left[node] = right[node] = 1;} else {if(L == R){len[node] = 0;cnt[node] = 0;left[node] = right[node] = 0;}else{CHD;len[node] = len[lc]+len[rc];cnt[node] = cnt[lc]+cnt[rc]-(left[rc]&right[lc]);left[node] = left[lc];right[node] = right[rc];}}}void update(int from,int to,int val,int node,int L,int R){if(from <= L && R <= to){cov[node] += val;} else {MID;CHD;if(from <= mid) update(from,to,val,lc,L,mid);else    maintain(lc,L,mid);if(to > mid) update(from,to,val,rc,mid+1,R);else    maintain(rc,mid+1,R);}maintain(node,L,R);}int solve(){int ans = 0,ans2 = 0;//竖直和水平方向的周长int n = vs.size()-1;int last = 0;rep(i,0,line.size()-1){int x = line[i].x;int f = line[i].y1+1;int t = line[i].y2;if(i > 0){int tmp = cnt[1] * (x-line[i-1].x);ans2 += tmp*2;}update(f,t,line[i].flag,1,1,n);ans += abs(last-len[1]);//竖直方向长度变化last = len[1];}return ans+ans2;}
}tree;void pre(){sort(vs.begin(),vs.end());vs.erase(unique(vs.begin(),vs.end()),vs.end());rep(i,0,line.size()-1){line[i].y1 = lower_bound(vs.begin(),vs.end(),line[i].y1) - vs.begin();line[i].y2 = lower_bound(vs.begin(),vs.end(),line[i].y2) - vs.begin();}sort(line.begin(),line.end());
}int main(){int n;scanf("%d",&n);if(n == 0){printf("0\n");return 0;}line.reserve(n<<1);vs.reserve(n<<1);rep(i,1,n){int x1,y1,x2,y2;scanf("%d%d%d%d",&x1,&y1,&x2,&y2);vs.push_back(y1);vs.push_back(y2);line.push_back(Node(x1,y1,y2,1));line.push_back(Node(x2,y1,y2,-1));}pre();int ans = tree.solve();printf("%d\n",ans);return 0;
}

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