2015-2016 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)

题目链接：http://codeforces.com/contest/589

I题：水题签到。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);

int cnt[maxn];
int n,k,a;

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>n>>k){
        MS0(cnt);
        REP(i,1,n) scanf("%d",&a),cnt[a]++;
        int x=n/k,ans=0;
        REP(i,1,k){
            ans+=abs(cnt[i]-x);
        }
        cout<<ans/2<<endl;
    }
    return 0;
}

View Code

J题：水题，模拟。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=30;
const int INF=(1<<29);

int dir[maxn*maxn*maxn];
int n,m;
bool vis[maxn][maxn][4];
char ch[maxn][maxn];
int sx,sy,sdir;
int ans;
bool vi[maxn][maxn];

void Init()
{
    dir['L']=0;
    dir['U']=1;
    dir['R']=2;
    dir['D']=3;
}

bool can(int x,int y,int d)
{
    if(x<1||x>n||y<1||y>m) return 0;
    return ch[x][y]=='.';
}

void dfs(int x,int y,int d)
{
    if(vis[x][y][d]) return;
    vis[x][y][d]=1;
    if(!vi[x][y]) ans++,vi[x][y]=1;
    //cout<<x<<" "<<y<<" "<<ans<<" "<<endl;
    if(d==0){
        if(can(x,y-1,d)) dfs(x,y-1,d);
        else dfs(x,y,(d+1)%4);
    }
    else if(d==1){
        if(can(x-1,y,d)) dfs(x-1,y,d);
        else dfs(x,y,(d+1)%4);
    }
    else if(d==2){
        if(can(x,y+1,d)) dfs(x,y+1,d);
        else dfs(x,y,(d+1)%4);
    }
    else{
        if(can(x+1,y,d)) dfs(x+1,y,d);
        else dfs(x,y,(d+1)%4);
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    Init();
    while(cin>>n>>m){
        MS0(vis);
        MS0(vi);
        REP(i,1,n){
            REP(j,1,m){
                cin>>ch[i][j];
                if(isupper(ch[i][j])){
                    sx=i,sy=j;
                    sdir=dir[ch[i][j]];
                    ch[i][j]='.';
                }
            }
        }
        ans=0;
        dfs(sx,sy,sdir);
        cout<<ans<<endl;
    }
    return 0;
}

View Code

B题：贪心水题。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);

int n;
struct Node
{
    ll a,b;
};
Node rec[maxn];
vector<ll> w;

bool cmp1(Node A,Node B)
{
    if(A.a<B.a) return 1;
    if(A.a==B.a) return A.b<B.b;
    return 0;
}

int sb()
{
    int m=0;
    int sz=(int)w.size();
    for(int i=0;i<w.size();i++){
        if(w[i]*(sz-i)>=w[m]*(sz-m)) m=i;
    }
    return m;
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>n){
        w.clear();
        REP(i,1,n){
            scanf("%I64d%I64d",&rec[i].a,&rec[i].b);
            if(rec[i].a<rec[i].b) swap(rec[i].a,rec[i].b);
            w.push_back(rec[i].b);
        }
        sort(rec+1,rec+n+1,cmp1);
        sort(w.begin(),w.end());
        ll ans=0,ansx,ansy;
        REP(i,1,n){
            int p=sb();
            ll x=rec[i].a,y=w[p];
            ll cnt=(int)w.size()-p;
            if(x*y*cnt>ans){
                ans=x*y*cnt;
                ansx=x;ansy=y;
            }
            vector<ll>::iterator it=find(w.begin(),w.end(),rec[i].b);
            w.erase(it);
        }
        cout<<ans<<endl;
        cout<<ansx<<" "<<ansy<<endl;
    }
    return 0;
}

View Code

F题：二分+贪心+线段树。二分是二分时间t。贪心是给右端点排序以确定安排的优先级，分类讨论可证明贪心的正确性。线段树是区间更新和统计可用的时间。

#include<bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

using namespace std;

typedef long long ll;
const int maxn=100100;
const int INF=(1<<29);

int n;
struct Node
{
    int s,t;
    friend bool operator<(Node A,Node B)
    {
        if(A.t<B.t) return 1;
        if(A.t==B.t) return A.s<B.s;
        return 0;
    }
};Node dish[maxn];
struct SegTree
{
    int l,r;
    int cnt;
    int lazy;
    int id;
};SegTree T[maxn<<2];
const int N=100010;

void push_up(int rt)
{
    T[rt].cnt=T[rt<<1].cnt+T[rt<<1|1].cnt;
}

void push_down(int rt)
{
    if(T[rt].lazy){
        T[rt<<1].lazy=T[rt<<1|1].lazy=1;
        T[rt<<1].cnt=T[rt<<1|1].cnt=0;
        T[rt].lazy=0;
    }
}

void build(int l,int r,int rt)
{
    T[rt].l=l;T[rt].r=r;
    T[rt].lazy=0;
    if(l==r){
        T[rt].cnt=1;
        T[rt].id=l;
        return;
    }
    push_down(rt);
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    push_up(rt);
}

void update(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R){
        T[rt].lazy=1;
        T[rt].cnt=0;
        if(l==r) T[rt].lazy=0;
        return;
    }
    push_down(rt);
    int m=(l+r)>>1;
    if(L<=m) update(L,R,lson);
    if(R>m) update(L,R,rson);
    push_up(rt);
}

int query_cnt(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R) return T[rt].cnt;
    push_down(rt);
    int m=(l+r)>>1;
    int res=0;
    if(L<=m) res+=query_cnt(L,R,lson);
    if(R>m) res+=query_cnt(L,R,rson);
    return res;
}

int query_pos(int p,int l,int r)
{
    while(l<r){
        int m=(l+r)>>1;
        int lcnt=query_cnt(l,m,1,N,1);
        if(lcnt>=p) r=m;
        else p-=lcnt,l=m+1;
    }
    return l;
}

bool check(int t)
{
    build(1,N,1);
    REP(i,1,n){
        int L=dish[i].s,R=dish[i].t;
        int cnt=query_cnt(L,R,1,N,1);
        if(cnt<t) return 0;
        int a=query_pos(1,L,R),b=query_pos(t,L,R);
        update(a,b,1,N,1);
    }
    return 1;
}

int bin(int l,int r)
{
    int res=0;
    while(l<r){
        int m=(l+r)>>1;
        if(check(m)) l=m+1,res=max(res,m);
        else r=m;
    }
    return res;
}

int main()
{
   // freopen("in.txt","r",stdin);
    while(cin>>n){
        REP(i,1,n) scanf("%d%d",&dish[i].s,&dish[i].t),dish[i].s++;
        sort(dish+1,dish+n+1);
        cout<<bin(0,N)*n<<endl;
    }
    return 0;
}

View Code

来源：https://www.cnblogs.com/--560/p/4890362.html

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