#include #include #include using namespace std;

struct block { int value; int count; int begin; };

int main() { int n = 0; scanf("%d", &n);

vector<block> myblocks;
int intput = 0;
for (int i=1; i<=n; i++)
{
	scanf("%d", &intput);
	if (0 == myblocks.size())
	{
		block first_block;
		first_block.value = intput;
		first_block.count = 1;
		first_block.begin = i;
		myblocks.push_back(first_block);
	}
	else
	{
		block &cur_block = myblocks.back();

		if (intput == cur_block.value)
		{
			cur_block.count++;
		}
		else
		{
			block new_block;
			new_block.value = intput;
			new_block.begin = i;
			new_block.count = 1;
		    myblocks.push_back(new_block);
		}
	}
	
}

while (0 != myblocks.size())
{
    vector<block>::iterator iter = myblocks.begin();
    int cur_value = -1;
    while (iter != myblocks.end())
    {
        if (0==(*iter).count)
        {
            iter = myblocks.erase(iter);
            continue;
        }
        else
        {
            if (cur_value != (*iter).value)
            {
                cout << (*iter).begin << ' ';
                cur_value = (*iter).value;
                (*iter).begin++;
                (*iter).count--;
            }
            iter++;
        }
    }	
    cout << endl;
}

return 0;

}

debug了半小时,结果发现......

$\color {grey} 数组开得不够大!!!$

想引用一下曾经最近没开$long long$时写的:

不过数据范围$1≤n≤2×10^5$我数组开到$200001$评测只因你告诉我Standard answer longer than user output....

得了正题。

用$2$个动态数组,$tmp$是局部的,每次存删除后的水果序列,再赋值给最终的$ve$。

code:

#include <bits/stdc++.h>
using namespace std;

int n,a[300000],l[300000],r[300000];//报复性开大
vector <int> ve;
vector <int> tmp;

signed main()
{
	freopen("fruit.in","r",stdin);
	freopen("fruit.out","w",stdout);
    scanf("%d",&n);
    a[0] = a[n + 1] = -1;
	r[0] = 1;
	l[n + 1] = n;
    for(int i = 1;i <= n;i++)
	{
        scanf("%d",&a[i]);
        if(a[i] != a[i - 1]) ve.push_back(i);
        l[i] = i - 1;
		r[i] = i + 1;
    }
    while(r[0] != n + 1)
	{
		tmp.clear();
		for(int i = 0;i < ve.size();i++)
		{
            printf("%d ",ve[i]);
            int u = l[ve[i]];
			int v = r[ve[i]];
            r[u] = v;
			l[v] = u;
			if(a[ve[i]] != a[u] && a[ve[i]] == a[v]) tmp.push_back(v); 
		}
		ve = tmp;
		printf("\n");
    }
    return 0;
}