[考试总结]noip模拟38
作者：互联网
。。。。
\(boom\)
不知道怎么的 \(T1\) 上来我就给跳过了，然后就开始先干\(T3\)，感觉并不是很简单，但是也不是能说是很难。
然后我就突然想到了一种可以过掉一半数据的 \(dp\)，之后居然一下子就成功了。
大样例一测，过了？！
之后感觉还行，然后回头管 \(T1\)，但是却没有发现那个极其显然的双指针，然后我人无，还在 \(l\) 和 \(r\) 输出之前就特判 \(l\),\(r\)，然后惨烈。。。。
\(T2\) 本以为有一些部分分数，然而一分也没有。。。。。
a
我们如果固定左上的那一个端点，就会发现这个对于接下来的 \(n-i+1\) 行，它的合法范围其实都是有关联的。
如果我们使用莫队的思想 \((two\;pointers)\)，就会每一行均摊一个 \(m\)，之后就是 \(\mathcal O(n^2m)\)
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define asm(i,x) for(register signed i=head[x];i;i=edge[i].next)
namespace xin_io
{
#define debug cout<<"debug"<<endl
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
#define scanf ak = scanf
int ak;
}
#define int long long
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7;
namespace xin
{
#define calc(x1,y1,x2,y2) (he[x2][y2] - he[x2][y1-1] - he[x1-1][y2] + he[x1-1][y1-1])
int n,m;
int l,r,ans = 0;
char s[maxn];
int he[31][maxn/10];
int los[maxn],ros[maxn];
inline short main()
{
scanf("%lld%lld",&n,&m);
try(i,1,n)
{
scanf("%s",s+1);
try(j,1,m)
{
he[i][j] = he[i-1][j] + he[i][j-1] - he[i-1][j-1] + (s[j] == '1');//,cout<<he[i][j]<<' ';
// if(s[j] == '1') sp1 = 0;
}
}
scanf("%lld%lld",&l,&r);
if(l == 0 and r == n * m)
{
cout<<((n + 1) * n / 2) * ((m + 1) * m/ 2)<<endl;
return 0;
}
try(i,0,n-1)
{
memset(los,0,sizeof(int) * (n + 1));
memset(ros,0,sizeof(int) * (n + 1));
try(j,0,m-1)
{
try(x,i+1,n)
{
try(y,los[x],m)
{
if(he[x][y] - he[i][y] - he[x][j] + he[i][j] >= l)
{los[x] = y; break;}
if(y == m) los[x] = m + 1;
}
try(y,ros[x]+1,m)
{
if(he[x][y] - he[i][y] - he[x][j] + he[i][j] > r)
{ros[x] = y - 1; break;}
if(y == m) ros[x] = m;
}
ans += ros[x] - los[x] + 1;
}
}
}
cout<<ans<<endl;
return 0;
}
}
signed main() {return xin::main();}
b
题解写的多好
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define asm(i,x) for(register signed i=head[x];i;i=edge[i].next)
namespace xin_io
{
#define debug cout<<"debug"<<endl
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf; int ak; typedef long long ll; typedef unsigned long long ull;
class xin_stream{public:template<typename type>inline xin_stream &operator >> (type &x)
{
register type s = 0; register int f = 1; register char ch = gc();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return x = s * f,*this;
}}io;
}
#define int long long
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7,mod = inf; const ll llinf = 1e18+7;
namespace xin
{
int cnt[21][maxn/10],n,m,ans = 0,ton[21][maxn/10],s[maxn],maxx = -inf;
inline short main()
{
io >> n >> m;
try(i,1,n)try(j,1,m)
{
register int x; io >> x;
ton[i][x]++; maxx = std::max(maxx,x);
}
try(i,1,n) try(j,1,maxx) for(register int k=1;k*j<=maxx;++k)
cnt[i][j] += ton[i][j * k];
throw(j,maxx,1)
{
s[j] = 1; try(i,1,n) (s[j] *= (cnt[i][j] + 1) % mod) %= mod; s[j] --;
if(!s[j]) continue;
for(register int k=2;k * j <=maxx;++k) s[j] = (s[j] - s[k*j] + mod) % mod;
(ans += s[j] * j % mod) %= mod;
}
cout<<ans<<endl;
return 0;
}
}
signed main() {return xin::main();}
c
这个虽然没有想出正解点分治，但是也是暴力分数中最大的想法了。
就是我们对于每次询问，遍历整张图，找到 \(x\) 到 \(y\) 经过的点的路径。
之后因为每两个点之间的边不止一条，而我们又要找到最大的价值，发现贪心不可，便开始 \(dp\)
我们考虑 \(f_{i,j}\) 为处理到 \(i\) 点，然后并且选择了这个点的颜色为 \(j\)
转移显然:
\[f_{i,j} = max_{k \ne }(f_{i-1,k})+1\\ f_{i,j} = max_{k = j}(f_{i-1,k}) \]然后就是 \(\mathcal O(nq)\)，只能过前 \(5\) 点。。。
#include<bits/stdc++.h>
using std::cout; using std::endl;
#define try(i,a,b) for(register signed i=a;i<=b;++i)
#define throw(i,a,b) for(register signed i=a;i>=b;--i)
#define asm(i,x) for(register signed i=head[x];i;i=edge[i].next)
namespace xin_io
{
#define debug cout<<"debug"<<endl
#define sb(x) cout<<#x" = "<<x<<' '
#define jb(x) cout<<#x" = "<<x<<endl
#define gc() p1 == p2 and (p2 = (p1 = buf) + fread(buf,1,1<<20,stdin),p1 == p2) ? EOF : *p1++
char buf[1<<20],*p1 = buf,*p2 = buf; int ak; typedef long long ll; typedef unsigned long long ull;
class xin_stream{public:template<typename type>inline xin_stream &operator >> (type &x)
{
register type s = 0; register int f = 1; register char ch = gc();
while(!isdigit(ch)) {if(ch == '-') f = -1; ch = gc();}
while( isdigit(ch)) s = (s << 1) + (s << 3) + (ch xor 48),ch = gc(); return x = s * f,*this;
}}io;
}
using namespace xin_io; static const int maxn = 1e6+10,inf = 1e9+7; const ll llinf = 1e18+7;
namespace xin
{
class xin_edge{public:int next,ver,w;}edge[maxn];
int head[maxn],ced = 0;
inline void add(int x,int y,int z) {edge[++ced].ver = y;edge[ced].w = z; edge[ced].next = head[x]; head[x] = ced;}
int n,m,qnum;
bool vis[maxn];
int pre[maxn];
void dfs(int x,int goal)
{
vis[x] = 1;
if(x == goal) return;
asm(i,x)
{
register int y = edge[i].ver;
if(!vis[y])
{
pre[y] = x;
dfs(y,goal);
}
}
}
void outp(int x)
{
if(!x) return;
cout<<x<<' ';
outp(pre[x]);
}
std::vector<int>vec[501][501];
int f[501][10001];
int maxx = 0,ans = -inf;
void dp(int x,int i)
{
if(!x)
{
return ;
}
for(auto j : vec[x][pre[x]])
{
try(k,1,maxx)
if(k xor j)
f[i][j] = std::max(f[i-1][k] + 1,f[i][j]),ans = std::max(ans,f[i][j]);
else
f[i][j] = std::max(f[i-1][k],f[i][j]),ans = std::max(ans,f[i][j]);
}
dp(pre[x],i+1);
}
class xin_data
{
public:
int x,y,z;
}d[maxn];
int lisan[maxn],cnt;
bool sp1 = 1;
int top[maxn],hson[maxn],dep[maxn],fa[maxn],siz[maxn];
void dfs1(int x,int f)
{
siz[x] = 1; dep[x] = dep[f] + 1; fa[x] =f;
asm(i,x)
{
register int y = edge[i].ver;
if(y == f) continue;
dfs1(y,x);
siz[x] += siz[y];
if(siz[y] > siz[hson[x]]) hson[x] = y;
}
}
void dfs2(int x,int t)
{
top[x] = t;
if(hson[x]) dfs2(hson[x],t);
asm(i,x)
{
register int y = edge[i].ver;
if(y == hson[x] or y == fa[x]) continue;
dfs2(y,y);
}
}
inline int lca(int x,int y)
{
while(top[x] xor top[y])
{
if(dep[top[x]] < dep[top[y]]) std::swap(x,y);
x = fa[top[x]];
}
if(dep[x] > dep[y]) return y; return x;
}
inline short main()
{
io >> n >> m;
try(i,1,m)
{
io >> d[i].x >> d[i].y >> d[i].z;
lisan[++cnt] = d[i].z;
}
io >> qnum; if(!qnum) {return 0;}
std::sort(lisan+1,lisan+cnt+1);
maxx = std::unique(lisan+1,lisan+cnt+1) - (lisan + 1);
try(i,1,m)
{
d[i].z = std::lower_bound(lisan+1,lisan+maxx+1,d[i].z) - lisan;
add(d[i].x,d[i].y,d[i].z); add(d[i].y,d[i].x,d[i].z);
if(n <= 500 and vec[d[i].x][d[i].y].size()) sp1 = 0;
if(n <= 500 )vec[d[i].x][d[i].y].push_back(d[i].z),vec[d[i].y][d[i].x].push_back(d[i].z);
}
try(que,1,qnum)
{
register int x,y; io >> x >> y;
memset(vis,0,sizeof(bool) * (n + 1));
memset(pre,0,sizeof(int ) * (n + 1));
try(i,1,n) try(j,1,maxx) f[i][j] = 0;
ans = 0;
dfs(x,y);
// outp(y);cout<<endl;
dp(y,1);
cout<<ans<<endl;
}
return 0;
}
}
signed main() {return xin::main();}
标签：std,38,noip,int,register,try,ch,lisan,模拟 来源： https://www.cnblogs.com/NP2Z/p/15139761.html