poj_1836

来源：互联网发布：国家网络信息发布管理编辑：程序博客网时间：2024/04/29 20:37

源地址：http://poj.org/problem?id=1836

题目大意就是从一排身高迥异的士兵中剔除最少的人，使得队伍的样子如同如下这幅图：

用LIS就行。

先从头来一遍，再从末尾来一遍，最后枚举红蓝两个点，找出最长的不降子序列，也就是剔除的人最少。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<set>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m;#define Mod 1000000007#define N 510#define M 1000100double height[N*2];int dp1[N*2];int dp2[N*2];int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    while(sf(n)!=EOF){    for(int i=0;i<n;i++)    scanf("%lf",&height[i]);    memset(dp1,0,sizeof dp1);    memset(dp2,0,sizeof dp2);    dp1[0] = 1;    for(int i=1;i<n;i++){<span style="white-space:pre"></span>//从头来    dp1[i] = 1;    for(int j=i-1;j>=0;j--){    if(height[i]>height[j] && dp1[j]+1>dp1[i]){    dp1[i] = dp1[j]+1;    }    }    }    dp2[n-1] = 1;    for(int i=n-2;i>=0;i--){<span style="white-space:pre"></span>//从末尾来    dp2[i] = 1;    for(int j=i+1;j<n;j++){    if(height[i]>height[j] && dp2[j]+1>dp2[i]){    dp2[i] = dp2[j]+1;    }    }    }    int ans = Max(dp1[n-1],dp2[0]);    for(int i=0;i<n;i++){<span style="white-space:pre"></span>//枚举两个点    for(int j=i+1;j<n;j++){    if(dp1[i]+dp2[j]>ans)    ans = dp1[i]+dp2[j];    }    }    printf("%d\n",n-ans);    }return 0;}

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