[Poi2011]Meteors 题解

 1 #include<cstdio>
 2 #define ll long long
 3 const int M=300005;
 4 int n,m,i,k,now,a[M],p[M],s[M],e[M],id[M],di[M],nex[M],fir[M],ans[M];
 5 ll c[M];//now表示当前已经经历了几次事件
 6 
 7 int read()
 8 {
 9     int x=0; char ch=getchar();
10     while (ch<48 || ch>57) ch=getchar();
11     while (ch>47 && ch<58) x=(x<<1)+(x<<3)+ch-48,ch=getchar();
12     return x;
13 }
14 
15 void add(int x,int y) { for (;x<=m;x+=x&-x) c[x]+=y; }
16 ll sum(int x) { ll ans=0; for (;x;x^=x&-x) ans=ans+c[x]; return ans; }
17 
18 void solve(int L,int R,int l,int r)
19 {
20     if (L>R) return;
21     if (l==r) { for (;L<=R;++L) ans[id[L]]=l; return; }
22     int mid=l+r>>1,i,j,x=L-1,y=R+1; ll tot;
23     while (now<mid)
24         if (s[++now]<=e[now]) add(s[now],a[now]),add(e[now]+1,-a[now]);
25         else add(s[now],a[now]),add(1,a[now]),add(e[now]+1,-a[now]);
26     for (;now>mid;--now)
27         if (s[now]<=e[now]) add(s[now],-a[now]),add(e[now]+1,a[now]);
28         else add(s[now],-a[now]),add(1,-a[now]),add(e[now]+1,a[now]);
29     for (i=L;i<=R;++i)
30     {
31         for (j=fir[id[i]],tot=0;j && (tot=tot+sum(j))<p[id[i]];j=nex[j]);
32         if (tot<p[id[i]]) di[--y]=id[i]; else di[++x]=id[i];
33     }
34     for (i=L;i<=R;++i) id[i]=di[i];
35     solve(L,x,l,mid),solve(y,R,mid+1,r);
36 }
37 
38 int main()
39 {
40     n=read(),m=read();
41     for (i=1;i<=m;++i) nex[i]=fir[k=read()],fir[k]=i;
42     for (i=1;i<=n;++i) p[i]=read(),id[i]=i;
43     for (k=read()+1,i=1;i<k;++i) s[i]=read(),e[i]=read(),a[i]=read();
44     s[k]=1,e[k]=m,a[i]=1000000000,solve(1,n,1,k);
45     for (i=1;i<=n;++i)
46         if (ans[i]==k) puts("NIE");
47         else printf("%d\n",ans[i]);
48     return 0;
49 }