FatMouse’s Speed-HDU-動態規劃

FatMouse's Speed

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6929    Accepted Submission(s): 3028
Special Judge

Problem DescriptionFatMouse 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. 

InputInput 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.  

OutputYour 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

 

先按體重 或 速度排序。

之後就是求其中的最大單調序列的長度了，這個是個很常見的問題了。

#include <iostream>#include <algorithm>#include <stdio.h>#include <string.h>using namespace std;int opt[1001], trace[1001], ans[1001];struct Node{int w,p,index;};bool cmp(const Node & a, const Node & b){return a.w < b.w;}Node mice[1001];int main() {//freopen("in.txt", "r", stdin);int k = 0;while(cin >> mice[k].w >> mice[k].p){mice[k].index = k+1;k++;}memset(opt, 0 ,sizeof(opt));memset(trace, 0 ,sizeof(trace));sort(mice, mice+k, cmp);int tmax = 0,index;for(int i=1; i<k; i++){for(int j=0; j<i; j++){if(mice[i].p  < mice[j].p && mice[i].w > mice[j].w){if(opt[j] + 1 > opt[i]){opt[i] = opt[j] + 1;//trace[i] = mice[j].index;trace[i] = j;}}}if(opt[i] > tmax){tmax = opt[i];index = i;}}for(int i=tmax; i>=0; i--){ans[i] = mice[index].index;index = trace[index];}//for(int i=0; i<k; i++)//cout << opt[i] << " ";//cout << endl;//for(int i=0; i<k; i++)//cout << trace[i] << " ";//cout << endl;cout <<tmax+1 << endl;for(int i=0; i<=tmax; i++)cout << ans[i] << endl;return 0;}