VK Cup 2017 - Round 1 (CDE)

771C Bear and Tree Jumps

大意: 给定树,每步能走到距离不超过$k$的任意点,记$f(s,t)$为$s$到$t$的最少步数,求$\sum\limits_{s<t}f(s,t)$

对于路径$(u,v)$, 假设距离为$d$, 那么贡献为$\lceil\frac{d}{k}\rceil=\frac{d+(-d\text{%}k+k)\text{%}k}{k}$

也就是说枚举每条边的贡献算出总路径长, 再$O(nk^2)$的$dp$求出多余部分, 最后除以$k$即为答案

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head

const int N = 1e6+;

int n, k;
ll ans, dp[N][], sz[N];
vector<int> g[N];
void dfs(int x, int fa) {
    dp[x][] = sz[x] = ;
    for (int y:g[x]) if (y!=fa) {
        dfs(y,x);
        sz[x] += sz[y];
        REP(i,,k-) REP(j,,k-) {
            ll t = (-i-j-)%k;
            if (t<) t += k;
            ans += t*dp[x][i]*dp[y][j];
        }
        REP(i,,k-) dp[x][(i+)%k]+=dp[y][i];
    }
    ans += sz[x]*(n-sz[x]);
}

int main() {
    scanf("%d%d", &n, &k);
    REP(i,,n) {
        int u, v;
        scanf("%d%d", &u, &v);
        g[u].pb(v),g[v].pb(u);
    }
    dfs(,);
    printf("%lld\n",ans/k);
}

771D Bear and Company

大意: 给定字符串, 每次操作任选两个相邻字符交换, 求最少操作次数使得不含子串"VK".

$dp$好题, 记${dp}_{v,k,x,0/1}$表示前$v$个'V',前$k$个'K',前$x$个其余字符,最后一个字符是否是'V'的最少操作次数, 然后暴力枚举字符, 算出排在它前面的字符数转移即可. 数据范围比较小可以直接$O(n^4)$, 也可以预处理一下达到$O(n^3)$

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head

const int N = ;
int n, dp[N][N][N][];
int a[][N], c[];
char s[N];
void chkmin(int &a, int b) {a>b?a=b:;};

int main() {
    scanf("%d%s", &n, s+);
    REP(i,,n) {
        if (s[i]=='V') a[][++c[]] = i;
        else if (s[i]=='K') a[][++c[]] = i;
        else s[i] = 'X', a[][++c[]] = i;
    }
    memset(dp,0x3f,sizeof dp);
    dp[][][][] = ;
    REP(v,,c[]) REP(k,,c[]) REP(x,,c[]) REP(z,,) {
        auto calc = [&](int p) {
            int vv=v,kk=k,xx=x,tot=p;
            REP(i,,p) {
                if (s[i]=='V'&&vv) --tot,--vv;
                if (s[i]=='K'&&kk) --tot,--kk;
                if (s[i]=='X'&&xx) --tot,--xx;
            }
            return tot;
        };
        int &r = dp[v][k][x][z];
        if (r==INF) continue;
        if (v!=c[]) chkmin(dp[v+][k][x][],calc(a[][v+]-)+r);
        if (!z&&k!=c[]) chkmin(dp[v][k+][x][],calc(a[][k+]-)+r);
        if (x!=c[]) chkmin(dp[v][k][x+][],calc(a[][x+]-)+r);
    }
    int ans = INF;
    REP(i,,) chkmin(ans,dp[c[]][c[]][c[]][i]);
    printf("%d\n", ans);
}

771E Bear and Rectangle Strips

大意: 给定2*n棋盘, 求最多选出多少个不相交的和为零的矩形.

区间$dp$的话很容易做, 但是复杂度是$O(n^2)$的, 难以优化.

考虑另一种$dp$, 记${dp}_{i,j}$为第一行到第$i$个,第二行到$j$个时的最大值.

这样状态数同样是$O(n^2)$的, 但是注意到转移时可以贪心, 每个位置只向后取一个最小的矩形, 显然答案不会变差, 那么这样相当于维护一个轮廓线$dp$, 轮廓线状态数是$O(n)$的.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=%P;for (a%=P;n;a=a*a%P,n>>=)if(n&)r=r*a%P;return r;}
ll inv(ll x){return x<=?:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=;char p=getchar();while(p<''||p>'')p=getchar();while(p>=''&&p<='')x=x*+p-'',p=getchar();return x;}
//head

const int N = 1e6+;
int n, a[][N], pos[][N], ans[N];
ll s[][N];
map<ll,int> v[];
pii ext[][N];

void chkmax(int &a, int b) {
    if (a<b) a = b;
}
void chkmax(pii &a, pii b) {
    if (b.y>a.y||b.y==a.y&&b.x<a.x) a = b;
}

void upd(int x, int y, int s) {
    if (x<y) chkmax(ans[y],s),chkmax(ext[][x],pii(y,s));
    else chkmax(ans[x],s),chkmax(ext[][y],pii(x,s));
}

//第一行x,第二行y,之前的最优值为s,往后拓展答案
void extend(int x, int y, int s) {
    upd(x+,y,s), upd(x,y+,s);
    if (pos[][x]) upd(pos[][x],y,s+);
    if (pos[][y]) upd(x,pos[][y],s+);
    if (x==y&&pos[][x]) upd(pos[][x],pos[][x],s+);
}

int main() {
    scanf("%d", &n);
    REP(i,,) REP(j,,n) scanf("%d",a[i]+j);
    REP(i,,) PER(j,,n) {
        s[i][j] = s[i][j+]+a[i][j];
        s[][j] += s[i][j];
    }
    REP(i,,) PER(j,,n+) {
        pos[i][j] = v[i][s[i][j]];
        v[i][s[i][j]] = j;
    }
    REP(i,,n) {
        extend(i,i,ans[i]);
        if (ext[][i].y) extend(i,ext[][i].x,ext[][i].y);
        if (ext[][i].y) extend(ext[][i].x,i,ext[][i].y);
    }
    printf("%d\n", ans[n+]);
}