Wannafly挑战赛19

Wannafly挑战赛19

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A. 队列Q

需要支持把一个元素移到队首，把一个元素移到队尾，移到队首就直接放到队首前面那个位置，原位置标为0，队尾同理。

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
typedef long long ll;
const int N = 30000200;
using namespace std;
int n,m;
int q[N],hd,x,ed,P[N];
char s[1111];
void F(int x) {int p = P[x];swap(q[p],q[hd]);P[x] = hd;hd--;
}
void L(int x){int p=P[x];swap(q[p],q[ed]);P[x]=ed;ed++;
}
int main() {scanf("%d",&n);hd = 500000;ed = hd + n - 1;rep(i,hd,ed)scanf("%d",&q[i]),P[q[i]]=i;hd--;ed++;scanf("%d",&m);rep(ti,1,m){scanf(" %s %d",s,&x);if(s[0]=='F'){F(x);}else {L(x);}}int f=0;rep(i,hd,ed)if(q[i]){if(f)printf(" ");f=1;printf("%d",q[i]);}puts("");return 0;
}

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B. 矩阵

带限制的最大子矩阵，首先是与不带限制的最大子矩阵一样，最上边的边和底边，然后做最大子段和。带上了0的限制和长度限制，于是写了个双指针，就过了。（讲道理感觉双指针有点假。。。
ps: 好像真的是假算法，回头补下单调栈做法。

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
typedef long long ll;
const int N = 555;
const ll inf = 1000000000000000LL;
using namespace std;
int R,C,X,Y,Z;
ll mp[N][N], sum[N][N], sum0[N][N], a[N], b[N];
int ck(int l,int r,ll x,ll y) {if(r-l+1 <= Y && l<=r && r<=C && x<=Z) return 1;return 0;
}
ll solve() {ll res=0,ans=a[1],ans0=b[1];rep(i,1,C)res=max(res,a[i]);int l = 1, r = 1;while(l<=r&&r<=C) {if(ck(l,r,ans0,ans))res = max(res,ans);while(ck(l,r+1,ans0+b[r+1],ans))++r,ans+=a[r],ans0+=b[r],res=max(res,ans);if(ck(l,r,ans0,ans))res = max(res,ans);ans -= a[l];ans0 -= b[l];++l;while(a[l]<=0&&l<=C)ans0-=b[l],ans-=a[l],++l;while(l>r)++r,ans0+=b[r],ans+=a[r];if(ck(l,r,ans0,ans))res = max(res,ans);}return res;
}
int main() {scanf("%d %d %d %d %d",&R,&C,&X,&Y,&Z);rep(i,1,R)rep(j,1,C)scanf("%lld",&mp[i][j]);rep(i,1,R)rep(j,1,C){sum[i][j] = sum[i-1][j] + mp[i][j];if(mp[i][j]==0) sum0[i][j] = sum0[i-1][j] + 1;else sum0[i][j] = sum0[i-1][j];}ll ans = 0;rep(l,1,R)rep(r,l,min(R,l+X-1)){rep(k,1,C) a[k] = sum[r][k]-sum[l-1][k], b[k] = sum0[r][k]-sum0[l-1][k];ans = max(ans,solve());}printf("%lld\n",ans);return 0;
}

C. 多彩的树

预处理每种颜色状态下的路径数，但是这次统计的包含这个状态的所有子状态。因此考虑容斥，当当前状态的颜色数为奇数时，加奇数，减偶数，当为偶数时加偶数减奇数。
\(eg1: a1a2 = (a1a2 + a1 + a2) - a1 - a2\)
\(eg2: a1a2a3 = (a1a2a3 + a1a2 + a2a3 + a1a3 + a1 + a2 + a3) - (a1a2 + a1 + a2) - (a2a3 + a2 + a3) - (a1a3 + a1 + a3) + a1 + a2 + a3\)
没有系统学习过容斥，碰见类似容斥，最好举几个例子找规律

#include <cstdio>
#include <cstring>
#include <queue>
#define rep(i,a,b) for(int i=a;i<=b;++i)
typedef long long ll;
const int N = 50004;
const ll mod = 1e9 + 7;
using namespace std;
struct edge{int e,nxt;}E[N<<1];
int h[N],cc;
void add(int u,int v) {E[cc].e = v;E[cc].nxt = h[u];h[u]=cc;++cc;
}
int n,k,x,y,a[N],col[12],vis[N],cnt[1<<11];
ll ans[12],dp[(1<<12)],num[(1<<12)];
ll dfs(int s) {ll res = 1;vis[s] = 1;for(int i=h[s];~i;i=E[i].nxt) {int v = E[i].e;if(!vis[v]&&col[a[v]]) {res += dfs(v);}}return res;
}
int main() {scanf("%d%d",&n,&k);rep(i,1,n) scanf("%d",&a[i]);memset(h,-1,sizeof(h));rep(i,1,n-1)scanf("%d%d",&x,&y),add(x,y),add(y,x);rep(i,1,(1<<k)-1) {dp[i]=cnt[i]=0;rep(j,0,k-1){col[j+1]=0;if(i&(1<<j))col[j+1]=1,++cnt[i];}rep(j,1,n)vis[j]=0;rep(j,1,n)if(!vis[j]&&col[a[j]]){ll tmp = dfs(j);dp[i] += tmp*(tmp-1)/2;dp[i]%=mod;}}rep(i,1,(1<<k)-1){num[i]=0;for(int j=i;j>0;--j)if((j|i)==i){if((cnt[i]-cnt[j])&1)num[i]-=dp[j];elsenum[i]+=dp[j];num[i]%=mod;num[i]+=mod;num[i]%=mod;}ans[cnt[i]] += num[i]; ans[cnt[i]]%=mod;}ans[1]+=n;ans[1]%=mod;ll res = 0;for(int i = k; i >= 0; --i) {res = ((res*131LL%mod + ans[i]%mod)%mod)%mod;}printf("%lld\n",res);return 0;
}