HDOJ_1160：FatMouse's Speed 解题报告

先对输入的数据按重量作第一关键字、速度作第二关键字进行排序，注意实际排序函数中是先处理第二关键字。本题从前向后递推就好，不必反序。ppt里面有个描述个人觉得有一定误导性，“设f[i]为Mice[i]至Mice[n]最长的序列长度。” 我看来，f[i]是从Mice[1]开始、以Mice[i]为结尾时，最长的满足单调条件的序列长度。

if( m[j].w!=m[i].w && m[j].s>m[i].s && m[j].length+1>m[i].length)

    m[i].length = m[j].length+1;  //此为递推关系

// HDOJ_1160.cpp : FatMouse's Speed

#include <iostream>
#include <algorithm>
using namespace std;
struct MICE
{
	int w,s;
	int length,index;
	int pre;
};
bool cmp( MICE x, MICE y )
{
	if( x.w==y.w )
		return x.s > y.s;
	else
		return x.w < y.w;
}
int main()
{
	MICE m[1001];
	int i=1, j, maxLength=1, maxIndex=1;
	while( cin >> m[i].w >> m[i].s )
    {
		m[i].index = i;
		m[i].length = 1;
		m[i].pre = 0;
		i++;
    }
    //cout<<i<<endl;
	int n = i-1;	//记录老鼠数量
	sort(m,m+n,cmp);

	for( i=2; i<=n; i++ )
		for( j=1; j<i; j++ )
			if( m[j].w!=m[i].w && m[j].s>m[i].s && m[j].length+1>m[i].length){ 
				m[i].length = m[j].length+1;
				m[i].pre = j;
			}

	for( i=2; i<=n; i++ )
		if( maxLength<m[i].length ){
			maxLength = m[i].length;
			maxIndex = i;
		}

	cout << maxLength << endl;
	int *maxPath = new int[maxLength];
	maxPath[maxLength] = maxIndex;
	for( i=maxLength-1; i>0; i-- ){
		maxPath[i] = m[maxPath[i+1]].pre;
		if( maxPath[i]==0 )
			maxPath[i] = maxPath[i+1];
	}
	for( i=1; i<=maxLength; i++ )
		cout << m[maxPath[i]].index << endl;
    return 0;
}