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 m11, m22,…, mnn then it must be the case that
Wm[1m[1] < Wm[2m[2] < … < Wm[nm[n]
and
Sm[1m[1] > Sm[2m[2] > … > Sm[nm[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
题目大意是找到一个最多的老鼠序列，使得序列中的老鼠的体重满足递增，相应老鼠的速度满足递减。 这种要和两个因素排列有关的一般都是先排一个因素，再dp另一个因素 这就很简单了 for i 1~n for j 1~i-1 a[i]>a[j] dp[i]=max(dp[i],dp[j]+1) ; 再用一个pre的数组记录路径。记录最大的点，然后往回推。 int t=maxi; i=0; while(t!=0) { res[i++]=t; t=pre[t]; }
#include<stdio.h>#include<algorithm>using namespace std;#define MAXN 1000struct Node{ int w,s;//重量和速度 int index;//最初的序号，避免排序后乱掉顺序，后面需要输出的 }mouse[MAXN+10];bool cmp(Node a,Node b)//先按照w从小到大排序，再按照y从大到小排序 { if(a.w<b.w) return 1; else if(a.w==b.w&&a.s>b.s)return 1; else return 0;} int dp[MAXN+10];//dp[i]表示以第i个数据结尾的符合要求的子列长度int pre[MAXN+10];//记录i对应的上一个数据int res[MAXN+10];//存放最终结果下标 int main(){ //freopen("test.in","r",stdin); //freopen("test.out","w",stdout); int i=1,j; while(scanf("%d%d",&mouse[i].w,&mouse[i].s)!=EOF) { dp[i]=1; pre[i]=0; mouse[i].index=i; i++; } int n=i-1; sort(mouse+1,mouse+1+n,cmp); int maxlen=0;//最长序列长度 int maxi;//最长序列的最后一个数下标 dp[1]=1; for(i=1;i<=n;i++) { for(j=1;j<i;j++) if(mouse[i].w>mouse[j].w&&mouse[i].s<mouse[j].s&&dp[j]+1>dp[i]) { dp[i]=dp[j]+1; pre[i]=j; if(dp[i]>maxlen) { maxi=i; maxlen=dp[i]; } } } int t=maxi; i=0; while(t!=0) { res[i++]=t; t=pre[t]; } printf("%d/n",i); while(i>0) { i--; printf("%d/n",mouse[res[i]].index); } return 0; }新闻热点
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