The 2018 ACM-ICPC Asia Nanjing Regional Programming Contest

A. Adrien and Austin

大意: $n$个石子, 编号$1$到$n$, 两人轮流操作, 每次删除$1$到$k$个编号连续的石子, 不能操作则输, 求最后胜负情况.

删除一段后变成两堆, 可以用$sg$函数打表找规律

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    int n,k;
    cin>>n>>k;
    if (!n) return puts("Austin"),0;
    if (k!=1) return puts("Adrien"),0;
    return puts(n&1?"Adrien":"Austin");
}

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D. Country Meow

大意: 给定空间$n$个点, 求构造一个点到$n$个点距离和最小.

等价于求一个最小的球覆盖所有的点, 可以三分套三分

 

E. Eva and Euro coins

大意: 给定$01$串$s$和$t$, 每次操作在$s$中选择一段连续相同的长为$k$的区间翻转, 求$s$是否能变为$t$

观察$1$: 操作可逆. 观察$2$: 长度为$k$的连续相同区间可以任意移动.

注意特判$k=1$

#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+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 1e6+50;
int n,k;
char s[N],t[N];
pii a[N];

void trans(char s[], int n) {
    int top = 0;
    REP(i,1,n) {
        if (!top||a[top].x!=s[i]) a[++top]=pii(s[i],1);
        else if (++a[top].y==k) --top;
    }
    int cnt = 0;
    REP(i,1,top) while (a[i].y--) s[++cnt] = a[i].x;
    REP(i,cnt+1,n) s[i] = '0';
}

int main() {
    scanf("%d%d%s%s",&n,&k,s+1,t+1);
    if (k==1) return puts("Yes"),0;
    trans(s,n),trans(t,n);
    puts(strcmp(s+1,t+1)?"No":"Yes");
}

View Code

 

 

  

G. Pyramid

大意: 求$n$层等边三角形中等边三角形的个数.

打表然后插值求系数, 答案为$\frac{n}{4}+\frac{11n^2}{24}+\frac{n^3}{4}+\frac{n^4}{24}$

#include <iostream>
#include <set>
#include <queue>
#define x first
#define y second
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
using namespace std;
typedef pair<int,int> pii;
int n;
set<pii> s;

void dfs(int d, int x, int y) {
    if (d>n) return;
    if (s.count(pii(x,y))) return;
    s.insert(pii(x,y));
    dfs(d+1,x-1,y+1);
    dfs(d+1,x+1,y+1);
}
int dis(pii u, pii v) {
    return (u.x-v.x)*(u.x-v.x)+3*(u.y-v.y)*(u.y-v.y);
}
int main() {
    cin>>n;
    dfs(0,0,0);
    vector<pii> v(s.begin(),s.end());
    int sz = v.size(), ans = 0;
    REP(i,0,sz-1) REP(j,i+1,sz-1) REP(k,j+1,sz-1) {
        int q=dis(v[i],v[j]),w=dis(v[i],v[k]),e=dis(v[j],v[k]);
        if (q==w&&w==e) ++ans;
    }
    printf("%d\n",ans);
}

暴力代码

 

H. Huge Discount

大意: 给定一个长$n$的串$s$, 只含数字'0','1','2', 有$n$个商品, 第$i$个商品价格为s[i...n]. 对于一个商品, 每次操作选两个连续且不相同的数删除, 可以进行任意次操作. 求$n$个商品最小价格和.

 

 

I. Magic Potion

大意: $n$个英雄, $m$个怪物, 给定每个英雄能杀死的怪物集合, 每个英雄只能从中选一个杀死. 可以选出$k$个英雄多杀一次, 求最大能杀怪物数.

最大流水题, 开个虚拟结点, 源点向虚拟结点连$k$即可.

#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;
const int N = 1e6+10;
const int SS = N-3, S = N-2, T = N-1, INF = 0x3f3f3f3f;
int n, m, k;
struct edge {
    int to,w,next;
    edge(int to=0,int w=0,int next=0):to(to),w(w),next(next){}
} e[N];
int head[N], dep[N], vis[N], cur[N], cnt=1;
queue<int> Q;
int bfs() {
    REP(i,1,n+m) dep[i]=INF,vis[i]=0,cur[i]=head[i];
    REP(i,SS,T) dep[i]=INF,vis[i]=0,cur[i]=head[i];
    dep[S]=0,Q.push(S);
    while (Q.size()) {
        int u = Q.front(); Q.pop();
        for (int i=head[u]; i; i=e[i].next) {
            if (dep[e[i].to]>dep[u]+1&&e[i].w) {
                dep[e[i].to]=dep[u]+1;
                Q.push(e[i].to);
            }
        }
    }
    return dep[T]!=INF;
}
int dfs(int x, int w) {
    if (x==T) return w;
    int used = 0;
    for (int i=cur[x]; i; i=e[i].next) {
        cur[x] = i;
        if (dep[e[i].to]==dep[x]+1&&e[i].w) {
            int f = dfs(e[i].to,min(w-used,e[i].w));
            if (f) used+=f,e[i].w-=f,e[i^1].w+=f;
            if (used==w) break;
        }
    }
    return used;
}
int dinic() {
    int ans = 0;
    while (bfs()) ans+=dfs(S,INF);
    return ans;
}
void add(int u, int v, int w) {
    e[++cnt] = edge(v,w,head[u]);
    head[u] = cnt;
    e[++cnt] = edge(u,0,head[v]);
    head[v] = cnt;
} 


int main() {
    scanf("%d%d%d",&n,&m,&k);
    add(S,SS,k);
    REP(i,1,n) {
        add(S,i,1);
        add(SS,i,1);
        int t;
        scanf("%d",&t);
        while (t--) {
            int x;
            scanf("%d",&x);
            add(i,n+x,1);
        }
    }
    REP(i,1,m) add(n+i,T,1);
    printf("%d\n",dinic());
}

View Code