!HDU 1025 Constructing Roads In JGShining's Kingdom--DP--(LIS算法)
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题意:在马路两边分别有n个城市,给出期望的n条路用于连接两边的城市,但是要求路不能有交叉,求在期望的n条中路实际能保留下来的最大的条数
分析:这题很好
1.本题抽象出来的模型应该是求最长上升(不下降)子序列
2.LIS的 nlog(n)算法:
O(n^2) 的算法是dp[i]保留以i结尾的最长上升子序列的长度,令k=dp[i],O(nlog(n))算法是从k的角度出发,设d(k)为在长度为 k 的序列中的最小的位置,即:d(k)=min(a[i]),其中 f[i]=k,然后二分,每次看a[i]是否大于等于d(len),如果满足,更新d(len)=a[i],且len++,如果不满足,二分找到适合它的位置更新d(j),注意d(k)中保留的并不是符合题意的子序列!
没理解没关系,会用就行
代码:
#include<iostream>#include<cstdio>using namespace std;int n,dp[500005],a[500005];int LIS(){int low,up,len=1;dp[1]=a[1];for(int i=1;i<=n;i++){low=1,up=len;while(low<=up){int mid=(low+up)>>1;if(dp[mid]>=a[i]) up=mid-1;else low=mid+1;}dp[low]=a[i];if(low>len) len++;}return len;}int main(){int k=1;while(scanf("%d",&n)!=EOF){int m,r;for(int i=1;i<=n;i++){ scanf("%d%d",&m,&r); a[m]=r; } int ans=LIS(); printf("Case %d:\n",k); printf("My king, at most %d road",ans);if(ans>1) printf("s"); printf(" can be built.\n\n"); k++;}}
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