HDU

来源：互联网发布：华为紧急数据怎么处理编辑：程序博客网时间：2024/06/03 20:50

题目：1---n的排列，给定一个区间，找出一种分组方法，使得分的组数最少

思路：莫队算法，每当做一次修改，判断是否满足分组+1和-1的条件就好了

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define PI acos(-1.0)#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }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;}// 0x3f3f3f3f3f3f3f3f// 0x3f3f3f3fconst int maxn=1e5+50;struct Ask{    int L,R,id;}ask[maxn];int ans[maxn];int arr[maxn],num[maxn];//从1开始计数int n,m,siz,temp;//n个数，m个询问，块的大小bool cmp(Ask a,Ask b){    if(a.L/siz!=b.L/siz) return a.L/siz<b.L/siz;    else return a.R<b.R;}void sub(int pos){    int x=arr[pos];    num[x]=0;    if(num[x-1]==0&&num[x+1]==0) temp--;    if(num[x-1]==1&&num[x+1]==1) temp++;}void add(int pos){    int x=arr[pos];    num[x]=1;    if(num[x-1]==0&&num[x+1]==0) temp++;    if(num[x-1]==1&&num[x+1]==1) temp--;}void solve(){    temp=0;    mm(num,0);    int L=1,R=0;    for(int i=1;i<=m;i++){        while(R<ask[i].R){            R++;            add(R);        }        while(R>ask[i].R){            sub(R);            R--;        }        while(L<ask[i].L){            sub(L);            L++;        }        while(L>ask[i].L){            L--;            add(L);        }        ans[ask[i].id]=temp;    }}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)            scanf("%d",&arr[i]);        for(int i=1;i<=m;i++){            ask[i].id=i;            scanf("%d%d",&ask[i].L,&ask[i].R);        }        siz=(int)sqrt(n*1.0);        sort(ask+1,ask+m+1,cmp);        solve();        for(int i=1;i<=m;i++)            printf("%d\n",ans[i]);    }    return 0;}

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