題意：給出N個矩形，M次詢問

每次詢問給出R個，問這R個矩形圍成的面積

解題思路：對于每次詢問，做一次線段樹求面積的并操作。

每個節點保存的信息有l，r，cover，len分別表示該節點表示的區間[l,r],該區間被線段完全覆蓋的次數以及被線段覆蓋的長度。

#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std;const int maxn = 25; int n,m,edge[maxn],tmp[maxn]; int tmpsize,size; //size表示線段樹葉子節點個數struct Segment {int l,r,cover,len; }tree[maxn<<2]; struct Rectangle {int x1,x2,y1,y2; }rec[maxn]; struct Line {int x,y1,y2;int val;Line(){}Line(int _x,int _y1,int _y2,int _val){x = _x;y1 = _y1;y2 = _y2;val = _val;} }; vector<Line> vec;void build(int rt,int l,int r) {tree[rt].l = l, tree[rt].r = r;tree[rt].cover = tree[rt].len = 0;if(l + 1 == r || l == r) return;int mid = (l + r) >> 1;build(rt<<1,l,mid);build(rt<<1|1,mid,r); }void insert(int rt,int l,int r,int val) {if(l <= tree[rt].l && tree[rt].r <= r){tree[rt].cover += val;if(tree[rt].cover == 0)tree[rt].len = tree[rt<<1].len + tree[rt<<1|1].len;else tree[rt].len = edge[tree[rt].r] - edge[tree[rt].l];return;}int mid = (tree[rt].l + tree[rt].r) >> 1;if(l < mid) insert(rt<<1,l,r,val);if(mid < r) insert(rt<<1|1,l,r,val);if(tree[rt].cover == 0)tree[rt].len = tree[rt<<1].len + tree[rt<<1|1].len;else tree[rt].len = edge[tree[rt].r] - edge[tree[rt].l]; }bool cmp(Line a,Line b) {return a.x < b.x; }void DiscretData() {sort(tmp,tmp+tmpsize);size = 0;edge[++size] = tmp[0];for(int i = 1; i < tmpsize; i++)if(tmp[i] != tmp[i-1])edge[++size] = tmp[i]; }int bisearch(int val) {int l = 1, r = size, mid;while(l <= r){mid = (l + r) >> 1;if(edge[mid] == val) return mid;else if(edge[mid] < val)l = mid + 1;else r = mid - 1;} }int main() {int cas = 1;while(scanf("%d%d",&n,&m),n+m){for(int i = 1; i <= n; i++)scanf("%d%d%d%d",&rec[i].x1,&rec[i].y1,&rec[i].x2,&rec[i].y2);printf("Case %d:\n",cas++);while(m--){int r,q = 1;scanf("%d",&r);vec.clear();tmpsize = 0;for(int i = 1; i <= r; i++){int k;scanf("%d",&k);vec.push_back(Line(rec[k].x1,rec[k].y1,rec[k].y2,1));vec.push_back(Line(rec[k].x2,rec[k].y1,rec[k].y2,-1));tmp[tmpsize++] = rec[k].y1;tmp[tmpsize++] = rec[k].y2;}DiscretData();sort(vec.begin(),vec.end(),cmp);build(1,1,size);int ans = 0;for(int i = 0; i < vec.size(); i++){if(i > 0)ans += tree[1].len * (vec[i].x - vec[i-1].x);int l = bisearch(vec[i].y1);int r = bisearch(vec[i].y2);insert(1,l,r,vec[i].val);}printf("Query %d: %d\n",q++,ans);}}return 0; }

總結

以上是生活随笔為你收集整理的hdu 2461(线段树求面积并)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。