Codeforce 4D - Mysterious Present

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原题：

Peter decided to wish happy birthday to his friend from Australia and send him a card. To make his present more mysterious, he decided to make a chain. Chain here is such a sequence of envelopes A = {a1, a2, ..., an}, where the width and the height of the i-th envelope is strictly higher than the width and the height of the (i - 1)-th envelope respectively. Chain size is the number of envelopes in the chain.

Peter wants to make the chain of the maximum size from the envelopes he has, the chain should be such, that he'll be able to put a card into it. The card fits into the chain if its width and height is lower than the width and the height of the smallest envelope in the chain respectively. It's forbidden to turn the card and the envelopes.

Peter has very many envelopes and very little time, this hard task is entrusted to you.

Input

The first line contains integers n, w, h (1 ≤ n ≤ 5000, 1 ≤ w, h ≤ 106) — amount of envelopes Peter has, the card width and height respectively. Then there follow n lines, each of them contains two integer numbers wi and hi — width and height of the i-th envelope (1 ≤ wi, hi ≤ 106).

Output

In the first line print the maximum chain size. In the second line print the numbers of the envelopes (separated by space), forming the required chain, starting with the number of the smallest envelope. Remember, please, that the card should fit into the smallest envelope. If the chain of maximum size is not unique, print any of the answers.

If the card does not fit into any of the envelopes, print number 0 in the single line.

Sample test(s)
input
2 1 12 22 2
output
11 
input
3 3 35 412 119 8
output
31 3 2 

题目意思：

就是说，给你一个卡片的长度和宽度，然后再给你许多信封的大小，包括长度和宽度，然后信封的长度和宽度都要大于卡片的长度和宽度，然后就是寄出每个信封时，信封的大小也是要逐步递增的，然后就是要你考虑，最多能寄出多少封信

题目分析：

DP题，然后就是建立一个数组，来存储每个信封它之后能有多少个信封被容纳，也就是小于该信封，包括长宽，但是要大于卡片的大小，然后再输出这个序列就可以了

代码：

#include<iostream>#include<algorithm>#include<cstring>using namespace std;struct Envelope{int height;int width;};Envelope enList[5010];int result[5010];int ppp = 0;int resultIndex[5010];int envelopeNum = 0;int cardHeight = 0;int cardWidth = 0;int maxNum = 0;int maxIndex = 0;int getP(int m) {if(result[m]>0 || result[m] == -1) return result[m];if(enList[m].height>cardHeight && enList[m].width>cardWidth) result[m] = 1;else return -1;for (int i=1;i<=envelopeNum;i++) {if(i != m && enList[m].height>enList[i].height && enList[m].width>enList[i].width) {int op = getP(i) + 1;result[m] = max(result[m],op);}}if(maxNum<result[m]) {maxNum = result[m];maxIndex = m;}return result[m];}bool large(Envelope p1,Envelope p2) {if(p1.height>p2.height && p1.width>p2.width) return true;return false;}void print_s(int index){resultIndex[--ppp] = index;for (int i=1;i<=envelopeNum;i++) {if(result[index] == result[i]+1 && large(enList[index],enList[i])) print_s(i);}}int main(){while(cin>>envelopeNum>>cardWidth>>cardHeight) {memset(result,0,sizeof(result));memset(resultIndex,0,sizeof(resultIndex));maxNum = 0;maxIndex = 0;for (int i=1;i<=envelopeNum;i++) {cin>>enList[i].width>>enList[i].height;}for (int i=1;i<=envelopeNum;i++) if(result[i] == 0) getP(i);//cout<<"------  1 -------"<<endl;//for (int i=1;i<=envelopeNum;i++) cout<<result[i]<<" ";//cout<<"------  1 -------"<<endl;cout<<maxNum<<endl;if(!maxNum) continue;//cout<<maxIndex<<endl;ppp = maxNum;print_s(maxIndex);for (int i=0;i<maxNum;i++) {if(i==0) cout<<resultIndex[0];else cout<<" "<<resultIndex[i];}cout<<endl;}return 0;}

疑惑：

好的吧，其实代码没有AC，死在了第13组测试数据上了，我也不知道为什么，算法的应该是没错的，就是不知道细节是哪里错了，真是坑，求大神看到，帮忙分析下

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