101 - The Blocks Problem-优快云博客

本文深入探讨了C++中堆叠操作与堆积运算的实现细节，通过具体代码实例展示了如何在C++环境中进行高效的数据管理和操作。包括堆叠重置、移动元素、堆叠堆顶元素的堆顶操作、堆积操作等关键功能的实现，旨在提高编程技能和效率。

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#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

void reset(vector<int> blocks[],int blocksPos[],int a)
{
	while(blocks[blocksPos[a]].back()!=a)
	{
		blocks[blocks[blocksPos[a]].back()].push_back(blocks[blocksPos[a]].back());
		blocksPos[blocks[blocksPos[a]].back()]=blocks[blocksPos[a]].back();
		blocks[blocksPos[a]].pop_back();
	}
}

int main()
{
	vector<int> blocks[25];
	char command[5],option[5];
	int a,b,n,blocksPos[25];
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		blocks[i].push_back(i);
		blocksPos[i]=i;
	}
	while(scanf("%s",command))
	{
		if(command[0]=='q') break;
		scanf("%d%s%d",&a,option,&b);
		if(blocksPos[a]== blocksPos[b])
			continue;
		if(strcmp(command,"move")==0 && strcmp(option,"onto")==0)
		{
			reset(blocks,blocksPos,a);
			reset(blocks,blocksPos,b);
			blocks[blocksPos[b]].push_back(a);
			blocks[blocksPos[a]].pop_back();
			blocksPos[a]=blocksPos[b];
		}
		if(strcmp(command,"move")==0 && strcmp(option,"over")==0)
		{
			reset(blocks,blocksPos,a);
			blocks[blocksPos[b]].push_back(a);
			blocks[blocksPos[a]].pop_back();
			blocksPos[a]=blocksPos[b];
		}
		if(strcmp(command,"pile")==0 && strcmp(option,"onto")==0)
		{
			reset(blocks,blocksPos,b);
			vector<int>::iterator iter=blocks[blocksPos[a]].begin();
			for(;*iter!=a;iter++);
			blocks[blocksPos[b]].insert(blocks[blocksPos[b]].end(),iter,blocks[blocksPos[a]].end());
			blocks[blocksPos[a]].erase(iter,blocks[blocksPos[a]].end());
			vector<int>::reverse_iterator iter2=blocks[blocksPos[b]].rbegin();
			for(;*iter2!=a;iter2++)
				blocksPos[*iter2]=blocksPos[b];
			blocksPos[*iter2]=blocksPos[b];
		}
		if(strcmp(command,"pile")==0 && strcmp(option,"over")==0)
		{
			vector<int>::iterator iter=blocks[blocksPos[a]].begin();
			for(;*iter!=a;iter++);
			blocks[blocksPos[b]].insert(blocks[blocksPos[b]].end(),iter,blocks[blocksPos[a]].end());
			blocks[blocksPos[a]].erase(iter,blocks[blocksPos[a]].end());
			vector<int>::reverse_iterator iter2=blocks[blocksPos[b]].rbegin();
			for(;*iter2!=a;iter2++)
				blocksPos[*iter2]=blocksPos[b];
			blocksPos[*iter2]=blocksPos[b];
		}
	}
	for(int i=0;i<n;i++)
	{
		printf("%d:",i);
		for(int j=0;j<blocks[i].size();j++)
		{
			printf(" %d",blocks[i][j]);
		}
		printf("\n");
	}
	return 0;
}

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=37