题目大意:给出一些海报和贴在墙上的区间。问这些海报依照顺序贴完之后,最后能后看到多少种海报。
思路:区间的范围太大,然而最多仅仅会有10000张海报,所以要离散化。
之后用线段树随便搞搞就能过。
关键是离散化的方法,这个题我时隔半年才A掉,之前一直就TTT,我还以为是线段树写挂了。
当我觉得我自己的水平这样的水线段树已经基本写不挂的时候又写了这个题,竟然还是T。
后来我对照别人的代码,才发现是我的离散化写渣了。
以下附AC代码(79ms),这个离散化写的比較优雅。时间也非常快,以后就这么写了。
CODE(POJ 79ms):
#include <map> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 100010 #define LEFT (pos << 1) #define RIGHT (pos << 1|1) using namespace std; pair<int,int> interval[MAX]; pair<int,int *> src[MAX << 1]; int cases; int tree[MAX << 2]; int intervals; int cnt,t; bool color[MAX]; inline void Initialize() { t = cnt = 0; memset(color,false,sizeof(color)); memset(tree,0,sizeof(tree)); } inline void PushDown(int pos) { if(tree[pos]) { tree[LEFT] = tree[pos]; tree[RIGHT] = tree[pos]; tree[pos] = 0; } } inline void Modify(int l,int r,int x,int y,int pos,int c) { if(l == x && y == r) { tree[pos] = c; return ; } PushDown(pos); int mid = (l + r) >> 1; if(y <= mid) Modify(l,mid,x,y,LEFT,c); else if(x > mid) Modify(mid + 1,r,x,y,RIGHT,c); else { Modify(l,mid,x,mid,LEFT,c); Modify(mid + 1,r,mid + 1,y,RIGHT,c); } } void GetAns(int l,int r,int pos) { if(tree[pos]) { color[tree[pos]] = true; return ; } if(l == r) return ; int mid = (l + r) >> 1; GetAns(l,mid,LEFT); GetAns(mid + 1,r,RIGHT); } int main() { for(cin >> cases; cases; --cases) { scanf("%d",&intervals); Initialize(); for(int i = 1; i <= intervals; ++i) { scanf("%d%d",&interval[i].first,&interval[i].second); src[++cnt] = make_pair(interval[i].first,&interval[i].first); src[++cnt] = make_pair(interval[i].second,&interval[i].second); } sort(src + 1,src + cnt + 1); for(int i = 1; i <= cnt; ++i) { if(src[i].first != src[i - 1].first) ++t; *src[i].second = t; } for(int i = 1; i <= intervals; ++i) Modify(1,cnt,interval[i].first,interval[i].second,1,i); GetAns(1,cnt,1); int ans = 0; for(int i = 1; i <= intervals; ++i) ans += color[i]; printf("%d ",ans); } return 0; }
为了警醒后人,也为了警醒自己(以后再也不这么写离散化了)。我将我T的离散化代码也发上来,好孩子千万不要学习。
CODE(POJ TLE):
#include <map> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 1000010 #define LEFT (pos << 1) #define RIGHT (pos << 1|1) using namespace std; pair<int,int> interval[MAX]; map<int,int> G; int cases; int tree[MAX << 2]; int intervals,src[MAX << 1]; int cnt,t; bool color[MAX]; inline void Initialize() { G.clear(); t = cnt = 0; memset(color,false,sizeof(color)); memset(tree,0,sizeof(tree)); } inline void PushDown(int pos) { if(tree[pos]) { tree[LEFT] = tree[pos]; tree[RIGHT] = tree[pos]; tree[pos] = 0; } } inline void Modify(int l,int r,int x,int y,int pos,int c) { if(l == x && y == r) { tree[pos] = c; return ; } PushDown(pos); int mid = (l + r) >> 1; if(y <= mid) Modify(l,mid,x,y,LEFT,c); else if(x > mid) Modify(mid + 1,r,x,y,RIGHT,c); else { Modify(l,mid,x,mid,LEFT,c); Modify(mid + 1,r,mid + 1,y,RIGHT,c); } } void GetAns(int l,int r,int pos) { if(tree[pos]) { color[tree[pos]] = true; return ; } if(l == r) return ; int mid = (l + r) >> 1; GetAns(l,mid,LEFT); GetAns(mid + 1,r,RIGHT); } int main() { for(cin >> cases; cases; --cases) { scanf("%d",&intervals); Initialize(); for(int i = 1; i <= intervals; ++i) { scanf("%d%d",&interval[i].first,&interval[i].second); src[++cnt] = interval[i].first; src[++cnt] = interval[i].second; } sort(src + 1,src + cnt + 1); G[src[1]] = ++t; for(int i = 2; i <= cnt; ++i) if(src[i] != src[i - 1]) G[src[i]] = ++t; for(int i = 1; i <= intervals; ++i) Modify(1,cnt,G[interval[i].first],G[interval[i].second],1,i); GetAns(1,cnt,1); int ans = 0; for(int i = 1; i <= intervals; ++i) ans += color[i]; printf("%d ",ans); } return 0; }
后来我和小伙伴门一起研究了这样的离散化方法,正常来说是不会T的,尽管多带了一个log。可是这个题是多组数据啊。坑啊,每一组数据就要clear一下map。据老师说map的clear是一个一个删的。。
。于是机制的我就每次new一个map,这样就不用clear了。结果交上去果然A了。
CODE(POJ 344ms):
#include <map> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define MAX 1000010 #define LEFT (pos << 1) #define RIGHT (pos << 1|1) using namespace std; pair<int,int> interval[MAX]; map<int,int> *G; int cases; int tree[MAX << 2]; int intervals,src[MAX << 1]; int cnt,t; bool color[MAX]; inline void Initialize() { G = new map<int,int>; t = cnt = 0; memset(color,false,sizeof(color)); memset(tree,0,sizeof(tree)); } inline void PushDown(int pos) { if(tree[pos]) { tree[LEFT] = tree[pos]; tree[RIGHT] = tree[pos]; tree[pos] = 0; } } inline void Modify(int l,int r,int x,int y,int pos,int c) { if(l == x && y == r) { tree[pos] = c; return ; } PushDown(pos); int mid = (l + r) >> 1; if(y <= mid) Modify(l,mid,x,y,LEFT,c); else if(x > mid) Modify(mid + 1,r,x,y,RIGHT,c); else { Modify(l,mid,x,mid,LEFT,c); Modify(mid + 1,r,mid + 1,y,RIGHT,c); } } void GetAns(int l,int r,int pos) { if(tree[pos]) { color[tree[pos]] = true; return ; } if(l == r) return ; int mid = (l + r) >> 1; GetAns(l,mid,LEFT); GetAns(mid + 1,r,RIGHT); } int main() { for(cin >> cases; cases; --cases) { scanf("%d",&intervals); Initialize(); for(int i = 1; i <= intervals; ++i) { scanf("%d%d",&interval[i].first,&interval[i].second); src[++cnt] = interval[i].first; src[++cnt] = interval[i].second; } sort(src + 1,src + cnt + 1); (*G)[src[1]] = ++t; for(int i = 2; i <= cnt; ++i) if(src[i] != src[i - 1]) (*G)[src[i]] = ++t; for(int i = 1; i <= intervals; ++i) Modify(1,cnt,(*G)[interval[i].first],(*G)[interval[i].second],1,i); GetAns(1,cnt,1); int ans = 0; for(int i = 1; i <= intervals; ++i) ans += color[i]; printf("%d ",ans); } return 0; }