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
",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];
            else
                num[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
",res);
    return 0;
}