HDU 1160 FatMouse's Speed DP题解

标签：使用   os   io   for   art   ar   amp   table   

本题就先排序老鼠的重量，然后查找老鼠的速度的最长递增子序列，不过因为需要按原来的标号输出，故此需要使用struct把三个信息打包起来。

查找最长递增子序列使用动态规划法，基本的一维动态规划法了。

记录路径：只需要记录后继标号，就可以逐个输出了。

#include <stdio.h>
#include <algorithm>
using namespace std;

const int MAX_N = 1005;

struct MouseSpeed
{
	int id, w, s;
	bool operator<(const MouseSpeed &ms) const
	{
		return w < ms.w;
	}
};

int N;
MouseSpeed msd[MAX_N];
int post[MAX_N], tbl[MAX_N];

int main()
{
	N = 0;
	while (~scanf("%d %d", &msd[N].w, &msd[N].s))
	{
		msd[N].id = N+1;
		++N;
	}

	sort(msd, msd+N);	//fullfill condition 1: weight increase
	fill(tbl, tbl+N, 1);//initialize dynamic table

	post[N-1] = N-1;
	for (int i = N-2; i >= 0; i--)
	{
		post[i] = i;	//as the print out terminate term
		for (int j = i+1; j < N; j++)
		{
			if (msd[i].s>msd[j].s && msd[i].w!=msd[j].w && tbl[i]<tbl[j]+1)
			{//strictly increase, so don't forget msd[i].w must < msd[j].w
				tbl[i] = tbl[j]+1;//update longest subsequence
				post[i] = j;//record the post
			}
		}
	}

	int id = 0, maxSeq = 0;
	for (int i = 0; i < N; i++)//find the max sequence starting point
	{
		if (maxSeq < tbl[i])
		{
			id = i;
			maxSeq = tbl[i];
		}
	}

	printf("%d\n", maxSeq);
	printf("%d\n", msd[id].id);
	while (id != post[id])
	{
		id = post[id];
		printf("%d\n", msd[id].id);//print out the original indices
	}

	return 0;
}

HDU 1160 FatMouse's Speed DP题解,布布扣,bubuko.com

HDU 1160 FatMouse's Speed DP题解

标签：使用   os   io   for   art   ar   amp   table   

原文地址：http://blog.csdn.net/kenden23/article/details/38275003