FatMouse's Speed
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7280 Accepted Submission(s): 3218
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.
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.
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 1300 6000 2100 500 2000 1000 4000 1100 3000 6000 2000 8000 1400 6000 1200 2000 1900
Sample Output
4 4 5 9 7
1.先将老鼠的speed递减排序，
2.求排过序数组的 weight 最长递增子序列
3求最长递增子序列时，每个元素的dp数组中递增时，都要记录它前面的递增元素下标，
4.根据最大的递增长度那个元素，递增找到这个递增子序列
#include<iostream>
#include<stdio.h>
#include<algorithm>
#define N 1002
using namespace std;
struct mouse{
int id,w,v;
}m[N];
struct fun{
int len;
int i;
int pre;
}f[N];
bool cmp(mouse a,mouse b){
if(a.v>b.v) return true;
return false;
}
bool cmp1(fun a,fun b){
if(a.len<b.len) return true;
return false;
}
int main()
{
int i,j,w,v,max,cnt = 1,k = 1;
while(scanf("%d%d",&w,&v)!=EOF){
m[k].id = cnt++;
m[k].w = w; m[k].v = v;
k++;
}
sort(m+1,m+cnt,cmp);
f[1].len = 1; f[1].i = m[1].id;
for(i=2;i<cnt;i++){
for(max =0,j=1; j<i; j++){
if(m[i].w>m[j].w && f[j].len>max){
max = f[j].len;
f[i].pre = j;
}
}
f[i].len = max+1; f[i].i = m[i].id;
}
int index;
for(max=1,i=1;i<cnt;i++)
if(max<f[i].len) {
max= f[i].len;
index = i;
}
k = 0;
int p[N];
while(f[index].pre!=0){
p[k++] = f[index].i;
index = f[index].pre;
}p[k] = f[index].i;
printf("%d\n",max);
for(i=k;i>=0;i--) {
printf("%d\n",p[i]);
}
return 0;
}