题意：给你俯视图，要求依次输出正视图中可以看到的建筑物
题解：任意相邻的x间属性相同，所以离散化。
坑：unique只能对数组用。下标易错
list不能找某元素的next。用了个很麻烦的处理
数组：
#define _CRT_SECURE_NO_WARNINGS #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<cstdio> #include<string> #include<stack> #include<ctime> #include<list> #include<set> #include<map> #include<queue> #include<vector> #include<sstream> #include<iostream> #include<functional> #include<algorithm> #include<memory.h> //#define INF 0x3f3f3f3f #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) #define rep(i,t,n) for(int i =(t);i<=(n);++i) #define per(i,n,t) for(int i =(n);i>=(t);--i) #define mp make_pair #define pb push_back #define mmm(a,b) memset(a,b,sizeof(a)) //std::ios::sync_with_stdio(false); using namespace std; typedef long long ll; typedef unsigned long long ull; void smain(); #define ONLINE_JUDGE int main() { //ios::sync_with_stdio(false); #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif smain(); #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms.", _end_time - _begin_time); #endif return 0; } const int maxn = 1e2 + 5; struct building { int id; double x, y, w, d, h; bool operator<(const building& rhs)const { return x < rhs.x || (x == rhs.x&&y < rhs.y); } }b[maxn]; int n; double x[maxn * 2]; bool cover(int i, double mx) { return b[i].x<mx&&b[i].x + b[i].w>mx; } bool visible(int i, double mx) { if (!cover(i, mx))return false; rep(k, 1, n) if (b[k].y < b[i].y&&b[k].h >= b[i].h&&cover(k, mx))return false; return true; } void Run() { } void smain() { int kase = 0; while (cin >> n) { if (n == 0)break; rep(i, 1, n) { cin >> b[i].x >> b[i].y >> b[i].w >> b[i].d >> b[i].h; x[i * 2] = b[i].x; x[i * 2 + 1] = b[i].x + b[i].w; b[i].id = i; } sort(b + 1, b + 1 + n); sort(x + 2, x + 2 + n * 2); int m = unique(x + 2, x + 2 + n * 2) - x-2; if (kase++)cout << endl; printf("For map #%d, the visible buildings are numbered as follows:\n%d", kase, b[1].id); rep(i, 2, n) { bool vis = false; rep(j, 2,m ) if (visible(i, (x[j] + x[j + 1]) / 2)) { vis = true; break; } if (vis)cout << ' ' << b[i].id; } cout << endl; } }
用list
#define _CRT_SECURE_NO_WARNINGS #include<cstring> #include<cctype> #include<cstdlib> #include<cmath> #include<cstdio> #include<string> #include<stack> #include<ctime> #include<list> #include<set> #include<map> #include<queue> #include<vector> #include<sstream> #include<iostream> #include<functional> #include<algorithm> #include<memory.h> //#define INF 0x3f3f3f3f #define eps 1e-6 #define pi acos(-1.0) #define e exp(1.0) #define rep(i,t,n) for(int i =(t);i<=(n);++i) #define per(i,n,t) for(int i =(n);i>=(t);--i) #define mp make_pair #define pb push_back #define mmm(a,b) memset(a,b,sizeof(a)) //std::ios::sync_with_stdio(false); using namespace std; typedef long long ll; typedef unsigned long long ull; void smain(); #define ONLINE_JUDGE int main() { //ios::sync_with_stdio(false); #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); long _begin_time = clock(); #endif smain(); #ifndef ONLINE_JUDGE long _end_time = clock(); printf("time = %ld ms.", _end_time - _begin_time); #endif return 0; } const int maxn = 1e2 + 5; struct building { int id; double x, y, w, d, h; bool operator<(const building& rhs)const { return x < rhs.x || (x == rhs.x&&y < rhs.y); } }b[maxn]; int n; double x[maxn * 2]; list<double> l; bool cover(int i, double mx) { return b[i].x<mx&&b[i].x + b[i].w>mx; } bool visible(int i, double mx) { if (!cover(i, mx))return false; rep(k, 1, n) if (b[k].y < b[i].y&&b[k].h >= b[i].h&&cover(k, mx))return false; return true; } void Run() { } void smain() { int kase = 0; while (cin >> n) { if (n == 0)break; rep(i, 1, n) { cin >> b[i].x >> b[i].y >> b[i].w >> b[i].d >> b[i].h; l.push_back(b[i].x), l.push_back(b[i].x + b[i].w); b[i].id = i; } sort(b + 1, b + 1 + n); l.sort(); l.unique(); if (kase++)cout << endl; printf("For map #%d, the visible buildings are numbered as follows:\n%d", kase, b[1].id); rep(i, 2, n) { bool vis = false; int last = l.front(); for (auto t : l) if (t != l.front() && visible(i, (t + last) / 2)) { vis = true; break; } else last = t; if (vis)cout << ' ' << b[i].id; } cout << endl; } }