| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12087 | Accepted: 3303 |
Description
Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago, from the moment I saw you smile, as you were walking out of the classroom and turned your head back, with the soft sunset glow shining on your rosy cheek, I knew, I knew that I was already drunk on you. Then, after several months’ observation and PRying, your grace and your wisdom, your attitude to life and your aspiration for future were all strongly impressed on my memory. You were the glamorous and sunny girl whom I always dream of to share the rest of my life with. Alas, actually you were far beyond my wildest dreams and I had no idea about how to bridge that gulf between you and me. So I schemed nothing but to wait, to wait for an appropriate opportunity. Till now — the arrival of graduation, I realize I am such an idiot that one should create the opportunity and seize it instead of just waiting. These days, having parted with friends, roommates and classmates one after another, I still cannot believe the fact that after waving hands, these familiar faces will soon vanish from our life and become no more than a memory. I will move out from school tomorrow. And you are planning to fly far far away, to pursue your future and fulfill your dreams. Perhaps we will not meet each other any more if without fate and luck. So tonight, I was wandering around your dormitory building hoping to meet you there by chance. But contradictorily, your appearance must quicken my heartbeat and my clumsy tongue might be not able to belch out a Word. I cannot remember how many times I have passed your dormitory building both in Zhuhai and Guangzhou, and each time aspired to see you appear in the balcony or your silhouette that cast on the window. I cannot remember how many times this idea comes to my mind: call her out to have dinner or at least a conversation. But each time, thinking of your Excellence and my commonness, the predominance of timidity over courage drove me leave silently. Graduation, means the end of life in university, the end of these glorious, romantic years. Your lovely smile which is my original incentive to work hard and this unrequited love will be both sealed as a memory in the deep of my heart and my mind. Graduation, also means a start of new life, a footprint on the way to bright prospect. I truly hope you will be happy everyday abroad and everything goes well. Meanwhile, I will try to get out from puerility and become more sophisticated. To pursue my own love and happiness here in reality will be my ideal I never desert. Farewell, my princess! If someday, somewhere, we have a chance to gather, even as gray-haired man and woman, at that time, I hope we can be good friends to share this memory proudly to relight the youthful and joyful emotions. If this chance never comes, I wish I were the stars in the sky and twinkling in your window, to bless you far away, as friends, to accompany you every night, sharing the sweet dreams or going through the nightmares together.
Input
There are several test cases in the input. The first line of each case contains 3 integers: n, W, H, indicating the number of stars, the horizontal length and the vertical height of the rectangle-shaped window. Then n lines follow, with 3 integers each: x, y, c, telling the location (x, y) and the brightness of each star. No two stars are on the same point. There are at least 1 and at most 10000 stars in the sky. 1<=W，H<=1000000, 0<=x，y<2^31.Output
For each test case, output the maximum brightness in a single line.Sample Input
3 5 41 2 32 3 26 3 13 5 41 2 32 3 25 3 1Sample Output
56題意：經典題，給你一個矩陣的大小，然后告訴你n個點的坐標，每個點都有一個值，你可以用矩陣來框住一些點，然后讓你求出你能框出的點的總和的最大值是多少。
思路：范圍很大，馬上想到離散化。但發現并不好做，完全不知道怎么繼續下手，因為你的矩陣大小不可能也離散化掉的。
這里需要換一個思維.
轉換一下，就能把其轉換為求線段區間的最大值 每個星星所能影響的范圍[(x,y),(x+w-1,y+h-1)]且有一權值 它們重合就表示 能被這個矩形框在一起，也就是說，只要求出重合的矩形的權值最大就行了。以x從小到大排序，y值離散化，投影到y軸上，那么對于每個星星的縱坐標，y，y+h-1就是每個星星可以影響到的矩形 然后x，x+w-1就是一個進入事件和一個出去事件，其所帶的值互為相反數. node[1].sum 保存當前的最大值 當所有的矩形都遍歷一遍 取其中的最大值就是ans.
#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int MAXN=1e4+7;int n,w,h;struct node{ long long x,y1,y2,val; bool Operator<(const node &a)const { if(x!=a.x)return x<a.x; return val>a.val;//這個地方不能少，不然所得的值可能比答案小的。 }}seg[MAXN<<1];struct tree{ int l,r; long long sum,add;}tree[MAXN<<3];long long y[MAXN<<1];void build_tree(int i,int l,int r){ tree[i].l=l; tree[i].r=r; tree[i].add=tree[i].sum=0; if(l==r)return ; int mid=(l+r)>>1; build_tree(i<<1,l,mid); build_tree(i<<1|1,mid+1,r);}void push_down(int i)//向下更新{ tree[i<<1].add+=tree[i].add; tree[i<<1].sum+=tree[i].add; tree[i<<1|1].add+=tree[i].add; tree[i<<1|1].sum+=tree[i].add; tree[i].add=0;}void updata(int i,long long l,long long r,long long val){ if(l==y[tree[i].l]&&r==y[tree[i].r]) { tree[i].sum+=val; tree[i].add+=val; return ; } if(tree[i].add)push_down(i); int mid=(tree[i].l+tree[i].r)>>1; if(r<=y[mid])updata(i<<1,l,r,val); else if(l>y[mid])updata(i<<1|1,l,r,val); else { updata(i<<1,l,y[mid],val); updata(i<<1|1,y[mid+1],r,val); } tree[i].sum=max(tree[i<<1].sum,tree[i<<1|1].sum);//向上更新}int main(){ int i; while(~scanf("%d%d%d",&n,&w,&h)) { for(i=0;i<n;++i) { scanf("%I64d%I64d%I64d",&seg[i].x,&seg[i].y1,&seg[i].val); seg[i].y2=seg[i].y1+h-1; seg[i+n]=seg[i]; seg[i+n].x=seg[i].x+w-1; seg[i+n].val=-seg[i].val; y[i*2+1]=seg[i].y1; y[i*2+2]=seg[i].y2; } sort(y+1,y+2*n+1); sort(seg,seg+2*n); int cnt=unique(y+1,y+2*n+1)-y-1;//去重函數,離散化基本就有它 build_tree(1,1,cnt); long long ans=0; for(i=0;i<n*2;++i) { updata(1,seg[i].y1,seg[i].y2,seg[i].val); ans=max(ans,tree[1].sum); } printf("%I64d/n",ans); } return 0;}
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