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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35053 Accepted Submission(s): 13880
题目链接:HDU 1257
不明觉厉的新定理,定理内容参考这篇博客:http://www.cppblog.com/jie414341055/archive/2010/05/28/116632.html,这个定理大部分情况下可以用贪心搞搞代替
但是有了这个定理解法会更高大上一点= =|||,求个最长严格上升子序列长度即可
链的最少划分数=反链的最长长度
代码:
#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;
int arr[N],d[N];
void init()
{
CLR(arr,0);
CLR(d,0);
}
int main(void)
{
int n,i,j;
while (~scanf("%d",&n))
{
for (i=1; i<=n; ++i)
scanf("%d",&arr[i]);
int len=1;
d[len]=arr[len];
for (i=2; i<=n; ++i)
{
if(d[len]<arr[i])
d[++len]=arr[i];
else
{
int pos=lower_bound(d,d+len,arr[i])-d;
d[pos]=arr[i];
}
}
printf("%d\n",len);
init();
}
return 0;
}
HDU 1257 最少拦截系统(Dilworth定理+LIS)
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原文地址:http://www.cnblogs.com/Blackops/p/5853923.html