http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=4107

題意：至多刪除一個(gè)點(diǎn)，使得破音數(shù)量最少

題解：前綴+后綴+枚舉?

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; int t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N],pre[N],suf[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%d",&t);while(t--){scanf("%d",&n);memset(pre,0,sizeof(pre));memset(suf,0,sizeof(suf));for(int i=1;i<=n;i++)scanf("%d",&a[i]);if(n<=2){cout<<0<<endl;continue;}for(int i=2;i<n;i++){pre[i]=pre[i-1]+(a[i-1]<a[i]&&a[i+1]<a[i]);}for(int i=n-1;i>1;i--){suf[i]=suf[i+1]+(a[i-1]<a[i]&&a[i+1]<a[i]);}ans=min(suf[2]-(a[1]<a[2]&&a[3]<a[2]),pre[n-1]-(a[n-2]<a[n-1]&&a[n]<a[n-1]));ans=min(ans,min(pre[n-3]+(a[n-3]<a[n-2]&&a[n]<a[n-2]),suf[4]+(a[1]<a[3]&&a[4]<a[3])));for(int i=3;i<=n-2;i++){//cout<<pre[i]<<" "<<suf[i]<<" "<<pre[i-2]+suf[i+2]+(a[i-2]<a[i-1]&&a[i+1]<a[i-1])+(a[i-1]<a[i+1]&&a[i+2]<a[i+1])<<endl;ans=min(ans,pre[i-2]+suf[i+2]+(a[i-2]<a[i-1]&&a[i+1]<a[i-1])+(a[i-1]<a[i+1]&&a[i+2]<a[i+1]));}cout<<min(ans,pre[n-1])<<endl;}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

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