chris garneau - relief 的中英文歌词

Posted 2023-04-12

参考技术A 歌词：
Always so still I never will be like you
And you never will strike me better
It’s always so clear, you never really hear at all
And I fear that you won’t get better

As when you are not being nice
You are not nice, you are not nice, you are not nice
I’d rather leave you alone
I’m gonna leave you alone

You’ve been quiet for so long
Something was wrong
You never said it a word
So I know that you won’t get better

That’s your fifth drink
Don’t you think that’s a lot?
Considering we’ve only been here for a little while now

When you are not being nice
you are not nice, you are not nice, you are not nice
I’d rather leave you alone
I’m gonna leave you alone

You are always so still
I never will be like you
And you never will strike me a better at all

Sometimes you are nice
But when you are not nice, you are not nice, you are not nice
I’d rather leave you alone
I’m gonna leave you alone

总是这么安静，我决不会像你这样
你也不会让我觉得好受点
一切如此清晰，你从未真正听过
我担心你不会改变
因为当你不友好的时候
你不友好，你不友好，你不友好
我宁愿让你独自一人
我会让你独自一人
你安静了好久
出了问题
你一言不发
于是我知道你不会改变
那是你的第五杯了
你不觉得太多了吗
我们只在这儿呆了一会儿
当你不友好的时候
你不友好，你不友好，你不友好
我宁愿让你独自一人
我会让你独自一人
你总是这么安静
我决不会像你这样
你也不会让我觉得好受点
有时候你很友好
但是当你不友好时，你不友好，你不友好
我宁愿让你独自一人
我会让你独自一人

bzoj5029 Relief grain

题目链接

树剖+线段树

将区间修改转化为单点修改，因为如果按DFS序进行修改，那么一定会对DFS序更大的点造成影响

  1 #include<iostream>
  2 #include<vector>
  3 #include<cstdio>
  4 #include<cstdlib>
  5 #include<string>
  6 #include<cstring>
  7 #include<cmath>
  8 #include<algorithm>
  9 #define re(i,l,r) for(int i=(l);i<=(r);i++)
 10 using namespace std;
 11 template <typename Q>
 12 void inin(Q &ret)
 13 {
 14     ret=0;int f=0;char ch=getchar();
 15     while(ch<‘0‘||ch>‘9‘){if(ch==‘0‘)f=1;ch=getchar();}
 16     while(ch>=‘0‘&&ch<=‘9‘)ret=(ret<<3)+(ret<<1)+ch-‘0‘,ch=getchar();
 17     ret=f?-ret:ret;
 18 }
 19 const int xxx=100010;
 20 int head[xxx],Next[xxx<<1],zhi[xxx<<1],n,m,dui[xxx],ans[xxx];
 21 int top[xxx],shen[xxx],fa[xxx],ed,dfn[xxx],son[xxx];
 22 vector<int>in[xxx],out[xxx];
 23 void add(int a,int b)
 24 {
 25     Next[++ed]=head[a],head[a]=ed,zhi[ed]=b;
 26     Next[++ed]=head[b],head[b]=ed,zhi[ed]=a;
 27 }
 28 int dfs(int x)
 29 {
 30     int ret=1,Max=0,temp;
 31     for(int i=head[x];i;i=Next[i])if(zhi[i]!=fa[x])
 32     {
 33         fa[zhi[i]]=x,shen[zhi[i]]=shen[x]+1;
 34         temp=dfs(zhi[i]);
 35         ret+=temp;
 36         if(Max<temp)Max=temp,son[x]=zhi[i];
 37     }
 38     return ret;
 39 }
 40 int tot;
 41 void dfs(int x,int t)
 42 {
 43     if(!x)return ;
 44     top[x]=t;dfn[x]=++tot,dui[tot]=x;
 45     dfs(son[x],t);
 46     for(int i=head[x];i;i=Next[i])if(zhi[i]!=fa[x]&&zhi[i]!=son[x])
 47         dfs(zhi[i],zhi[i]);
 48 }
 49 void change(int x,int y,int z)
 50 {
 51     while(top[x]!=top[y])
 52         if(shen[top[x]]>shen[top[y]])
 53             in[dfn[top[x]]].push_back(z),
 54             out[dfn[x]+1].push_back(z),
 55             x=fa[top[x]];
 56         else in[dfn[top[y]]].push_back(z),
 57             out[dfn[y]+1].push_back(z),
 58             y=fa[top[y]];
 59     if(shen[x]>shen[y])swap(x,y);
 60     in[dfn[x]].push_back(z),out[dfn[y]+1].push_back(z);
 61 }
 62 struct tree
 63 {
 64     int l,r,Max,o;
 65 }t[400040];
 66 void UP(int k)
 67 {
 68     int p1=k<<1,p2=p1|1;
 69     if(t[p1].Max>=t[p2].Max)t[k].Max=t[p1].Max,t[k].o=t[p1].o;
 70     else t[k].Max=t[p2].Max,t[k].o=t[p2].o;
 71 }
 72 void build(int k,int l,int r)
 73 {
 74     t[k].l=l,t[k].r=r;t[k].Max=0,t[k].o=l;
 75     if(l==r)return ;
 76     int mid=(l+r)>>1,p1=k<<1,p2=p1|1;
 77     build(p1,l,mid),build(p2,mid+1,r);
 78     UP(k);
 79 }
 80 void update(int k,int se,int x)
 81 {
 82     if(t[k].l==t[k].r){t[k].Max+=x;return ;}
 83     int mid=(t[k].l+t[k].r)>>1,p1=k<<1,p2=p1|1;
 84     if(se<=mid)update(p1,se,x);
 85     else update(p2,se,x);
 86     UP(k);
 87 }
 88 int main()
 89 {
 90     while(scanf("%d%d",&n,&m),n)
 91     {
 92         tot=0;
 93         ed=0;memset(head,0,sizeof head);
 94         memset(son,0,sizeof son);
 95         re(i,2,n)
 96         {
 97             int a,b;inin(a),inin(b);
 98             add(a,b);
 99         }
100         dfs(1);
101         dfs(1,1);int Max=0;
102 //        re(i,1,n)
103 //        {
104 //            printf("%d : ",i);
105 //            cout<<"in:";re(j,0,(int)in[i].size()-1)printf("%d ",in[i][j]);cout<<"\n";
106 //            cout<<"out:";re(j,0,(int)out[i].size()-1)printf("%d ",out[i][j]);cout<<"\n";
107 //        }
108         re(i,1,m)
109         {
110             int x,y,z;
111             inin(x),inin(y),inin(z);
112             change(x,y,z),Max=max(Max,z);
113         }
114         build(1,1,Max+1);
115         re(i,1,n)
116         {
117             re(j,0,(int)in[i].size()-1)
118                 update(1,in[i][j],1);
119             re(j,0,(int)out[i].size()-1)
120                 update(1,out[i][j],-1);
121             ans[dui[i]]=t[1].Max?t[1].o:0;
122         }
123         re(i,1,n)printf("%d\n",ans[i]);
124         re(i,1,n+1)in[i].clear(),out[i].clear();
125     }
126     return 0;
127 }