HDU 1160 FatMouse's Speed （dp+路径记录）

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FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16972    Accepted Submission(s): 7507
Special Judge

Problem Description

FatMouse believes that the fatter a mouse is, the faster it runs. To disprove this, you want to take the data on a collection of mice and put as large a subset of this data as possible into a sequence so that the weights are increasing, but the speeds are decreasing.

 

Input

Input contains data for a bunch of mice, one mouse per line, terminated by end of file.

The data for a particular mouse will consist of a pair of integers: the first representing its size in grams and the second representing its speed in centimeters per second. Both integers are between 1 and 10000. The data in each test case will contain information for at most 1000 mice.

Two mice may have the same weight, the same speed, or even the same weight and speed. 

 

Output

Your program should output a sequence of lines of data; the first line should contain a number n; the remaining n lines should each contain a single positive integer (each one representing a mouse). If these n integers are m[1], m[2],..., m[n] then it must be the case that 

W[m[1]] < W[m[2]] < ... < W[m[n]]

and 

S[m[1]] > S[m[2]] > ... > S[m[n]]

In order for the answer to be correct, n should be as large as possible.
All inequalities are strict: weights must be strictly increasing, and speeds must be strictly decreasing. There may be many correct outputs for a given input, your program only needs to find one. 

 

Sample Input

6008 13006000 2100500 20001000 40001100 30006000 20008000 14006000 12002000 1900

 

Sample Output

44597

题意是让求体重严格递增，速度严格递减的老鼠，并且输出老鼠的位置，表示没看到Special Judge，一直调不出示例结果，后来觉得自己很对，提交竟然A到。

可以按照体重递增排序，用ｄｐ[i]数组表示前i个老鼠中，有多少个是符合题目要求的，用path数组记录当前数组的前一个的值，使用结构体记录老鼠的初始位置。path数组初始化为每一个老鼠都只能到达本身，最后倒序输出。典型的dp模板题，就是多了路径记录。

代码实现：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;const int maxn=1e5+5;struct mouse{int weight;int feed;    //英语不好，不要在意细节，好像应该是speed int count;}p[maxn];int cmp(mouse a,mouse b){if(a.weight==b.weight)return a.feed<b.feed;return a.weight<b.weight;}int dp[maxn],path[maxn],ans[maxn];int main(){int i=0,j,k=0;while(cin>>p[i].weight){cin>>p[i].feed;p[i].count=++k;i++;}memset(dp,0,sizeof(dp));sort(p,p+i,cmp);int n=i;for(i=0;i<n;i++)path[i]=i;int maxx=-1,index=0;for(i=0;i<n;i++){dp[i]=1;for(j=0;j<i;j++){if(p[i].weight>p[j].weight&&p[i].feed<p[j].feed&&dp[i]<dp[j]+1){dp[i]=dp[j]+1;path[i]=j;if(dp[i]>maxx){maxx=dp[i];index=i;}}}}cout<<maxx<<endl;k=0;while(path[index]!=index){ans[k++]=p[index].count;index=path[index];}ans[k++]=p[index].count;for(i=k-1;i>=0;i--)cout<<ans[i]<<endl;return 0;}