Mayor's posters POJ2528 ACM算法设计
题目链接:http://poj.org/problem?id=2528
大致题意:
有一面墙,被等分为1千万份,一份的宽度为一个单位宽度。现在往墙上贴N张海报,每张海报的宽度是任意的,但是必定是单位宽度的整数倍,且<=1QW。后贴的海报若与先贴的海报有交集,后贴的海报必定会全部或局部覆盖先贴的海报。现在给出每张海报所贴的位置(左端位置和右端位置),问张贴完N张海报后,还能看见多少张海报
注意:
这面墙被分为1千万份,从1开始对每份命名,就是说,如果海报的起、终为5 5 ,则海报占据了第5个单元格,若为5 6,则占据了第5和第6个单元格。
错误的离散化:
如这组数据
1
3
1 10
1 3
6 10
不离散的话,正确答案是3,错误离散后答案成了2.
下面是没AC的代码,纪念逝去的10+个小时..
1 #include <iostream> 2 #include <algorithm> 3 #include <string.h> 4 //#include <fstream> 5 using namespace std; 6 7 struct LineTreeNode 8 { 9 int l,r,col; 10 }; 11 const unsigned int MAXnumber = (unsigned int)1e7; 12 const int MAXposters = 10000; 13 unsigned int posters[MAXposters+1][2]; 14 unsigned int mapping[MAXnumber+1]; 15 unsigned int posterPoints[MAXposters*2+1]; 16 bool IsAdded[MAXposters+1]; 17 LineTreeNode LT[MAXposters*2*4+1]; 18 int posterCount; 19 unsigned int lineTreeLength; 20 int ans; 21 22 void creatTree(const int& l,const int& r,const int& p) 23 { 24 LT[p].l = l; 25 LT[p].r = r; 26 if(l == r) 27 return; 28 int mid = (l + r) >> 1; 29 creatTree(l,mid,p * 2); 30 creatTree(mid + 1,r,p * 2 + 1); 31 } 32 void insertVal(const int& l,const int& r,const int& p,const int& col) 33 { 34 if(l > LT[p].r || r < LT[p].l) 35 return; 36 if(l <= LT[p].l && r >= LT[p].r) 37 { 38 LT[p].col = col; 39 return; 40 } 41 if(LT[p].col >= 0) 42 { 43 LT[p*2].col = LT[p*2 + 1].col = LT[p].col; 44 LT[p].col = -1; 45 } 46 insertVal(l,r,p*2,col); 47 insertVal(l,r,p*2+1,col); 48 } 49 void dfsSearch(const int& p) 50 { 51 if(LT[p].col > 0 && !IsAdded[LT[p].col]) 52 { 53 ans++; 54 IsAdded[LT[p].col] = true; 55 return; 56 } 57 if(LT[p].col == 0 || LT[p].l == LT[p].r) 58 return; 59 dfsSearch(p*2); 60 dfsSearch(p*2+1); 61 } 62 int main() 63 { 64 //ifstream cin("in.txt"); 65 int cases; 66 cin>>cases; 67 while (cases --) 68 { 69 cin>>posterCount; 70 //Init 71 ans = 0; 72 lineTreeLength = 0; 73 memset(IsAdded,false,sizeof(IsAdded)); 74 memset(mapping,0,sizeof(mapping)); 75 memset(posterPoints,0,sizeof(posterPoints)); 76 for(int i = 1;i < posterCount * 2 * 4 + 1;i ++) 77 LT[i].l = LT[i].r = LT[i].col = 0; 78 //Input 79 int p = 0; 80 for(int i = 1;i <= posterCount;i++) 81 { 82 cin>>posters[i][0]>>posters[i][1]; 83 if(!mapping[posters[i][0]]) 84 { 85 posterPoints[p++] = posters[i][0]; 86 mapping[posters[i][0]] = 1; 87 } 88 if(!mapping[posters[i][1]]) 89 { 90 posterPoints[p++] = posters[i][1]; 91 mapping[posters[i][1]] = 1; 92 } 93 } 94 //Sort 95 sort(posterPoints,posterPoints + p); 96 //Map 97 for(int i = 0;i < p;i ++) 98 { 99 mapping[posterPoints[i]] = ++lineTreeLength; 100 } 101 //Construct LineTree 102 creatTree(1,lineTreeLength,1); 103 //Place posters 104 for(int i = 1;i <= posterCount;i ++) 105 insertVal(mapping[posters[i][0]],mapping[posters[i][1]],1,i); 106 //How much can we see 107 dfsSearch(1); 108 //Print 109 cout<<ans<<endl; 110 } 111 }
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