2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

Posted 2020-10-06 BobHuang

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Bananas are the favoured food of monkeys.

In the forest, there is a Banana Company that provides bananas from different places.

The company has two lists.

The first list records the types of bananas preferred by different monkeys, and the second one records the types of bananas from different places.

Now, the supplier wants to know, whether a monkey can accept at least one type of bananas from a place.

Remenber that, there could be more than one types of bananas from a place, and there also could be more than one types of bananas of a monkey‘s preference.

Input Format

The first line contains an integer

For each test case, the first line contains two integers

In the each line of following

All integers of the input are less than or equal to

Output Format

For each test case, output all the pairs

These pairs should be outputted as ascending order. That is say that a pair of

If two pairs own the same

And there should be an empty line after each test case.

样例输入

样例输出

题目来源

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

#include<stdio.h>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        int m,n,i,x,y,mm=0,mi=100;
        vector<int>mon[500],pla[500],s;
        scanf("%d%d",&m,&n);
        for(i=0;i<m;i++){
            scanf("%d%d",&x,&y);
            mon[x].push_back(y);
            mm=max(mm,x);
            mi=min(mi,x);
        }
        for(i=0;i<n;i++){
            scanf("%d%d",&x,&y);
            pla[x].push_back(y);
        }
        int a[52],k=0,j,l;
        for(i=mi;i<=mm;i++){
            memset(a,0,sizeof a);
            for(j=0;j<mon[i].size();j++){
                for(l=0;l<pla[mon[i][j]].size();l++){
                    a[pla[mon[i][j]][l]]=1;
                }
            }
            for(j=0;j<51;j++){
                if(a[j])
                printf("%d %d\n",i,j);
            }
        }
        puts("");
    }
}

Coconut is Captain Gangplank‘s favourite fruit. That is why he needs to drink coconut juice from

On his next trip, he would pass through

His trip would begin in the

The journey from the $D_iD?i?? days.$

Initially, there is no coconut on his ship. Fortunately, he could get supply of $C_iC?i??coconuts from the ii-th city.$

Could you tell him, whether he could drink coconut juice every day during the trip no not?

Input Format

The first line contains an integer

For each test case the first line contains two integers

The second line contains $C_1, C_2, \cdots, C_NC?1??,C?2??,?,C?N??.$

The third line contains $D_1, D_2, \cdots, D_{N-1}D?1??,D?2??,?,D?N?1??.$

All integers in the input are less than

Output Format

For each case, output Yes if Captain Gangplank could drink coconut juice every day, and otherwise output No.

样例输入

样例输出

Yes
No

题目来源

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

#include<stdio.h>
int main()
{
    int a[1005],t,i,j,n,m,s,x,k;
    scanf("%d",&t);
    while(t--)
    {
        k=s=0;
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
        scanf("%d",&a[i]);
        for(i=0;i<n-1;i++)
        {
            s+=a[i];
            scanf("%d",&x);
            s-=x*m;
            if(s<0)
            {
                k=1;
                continue;
            }
        }
        if(k)puts("No");
        else puts("Yes");
    }
}

The numbers $t_i=\frac{1}{2}i(i+1)t?i??=?2??1??i(i+1), are called half-consecutive.$

For given $t_rt?r?? is square.$

Input Format

The input contains multiple test cases.

The first line of a multiple input is an integer

Each line contains an integer $N~(1\le N\le 10^{16})N (1≤N≤10?16??).$

Output Format

For each test case, output the case number first.

Then for given

If this half-consecutive number does not exist, output

样例输入

样例输出

Case #1: 1
Case #2: 8
Case #3: 49
Case #4: 288

题目来源

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

这几个数很少，打表找规律都行的

不过我读错了题意，这个锅我背

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
LL a[100]={1,
8,
49,
288,
1681,
9800,
57121,
332928,
1940449,
11309768,
65918161,
384199200,
2239277041,
13051463048,
76069501249,
443365544448,
2584123765441,
15061377048200,
87784138523761,
511643454094368,
2982076586042449,
17380816062160328};
int main()
{
    ios::sync_with_stdio(false);
    int T,k=1;
    scanf("%d",&T);
    while(T--)
    {
        LL n;
        scanf("%lld",&n);
        int pos=lower_bound(a,a+22,n)-a;
        printf("Case #%d: %lld\n",k++,a[pos]);
    }
    return 0;
}

问答

31.41%
1000ms
131072K

On the mysterious continent of Tamriel, there is a great empire founded by human.

To develope the trade, the East Empire Company is set up to transport goods from place to place.

Recently, the company wants to start their business in Solstheim, which is consists of

Luckily, there are already

All routes are one-way, and the $u_iu?i?? to v_iv?i??.$

Now, the company nominates you a particular job to plan some new routes to make sure that person and goods can be transported between any two islands.

Furthermore, because the neighboring regions are under attack by an increasing number of dragons, limited resources can be used to set up new routes.

So you should plan to build new routes as few as possible.

Input Format

The first line contains an integer

For each test case, the first line includes two integers $\leq 10000)N (N≤10000) and M~(M \leq 100000)M (M≤100000), as described above.$

After that there are $u_iu?i?? and v_iv?i??.$

Output Format

For each test case output one integer, represent the least number of routes required to new.

样例输入

样例输出

1
2

题目来源

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

这个求lca的，找个模板就能a

#include<bits/stdc++.h> 
#define N 100000+10  
using namespace std;  
vector<int>g[N];  
int low[N],dfn[N],dfs_clock, sccon[N],scc_cnt;  
bool instack[N];  
stack<int>s;  
int n,m;  
void tarjan(int u,int fa)  
{  
    int v;  
    low[u]=dfn[u]=++dfs_clock;  
    s.push(u);  
    instack[u]=true;  
    for(int i=0;i<g[u].size();i++)  
    {  
        v=g[u][i];  
        if(!dfn[v])  
        {  
            tarjan(v,u);  
            low[u]=min(low[u],low[v]);  
        }  
        else if(instack[v])  
            low[u]=min(low[u],dfn[v]);  
    }  
    if(low[u]==dfn[u])  
    {  
        scc_cnt++;  
        while(1)  
        {  
            v=s.top();  
            s.pop();  
            instack[v]=false;  
            sccon[v]=scc_cnt;  
            if(v==u)  
                break;  
        }  
    }  
}  
int in[N],out[N];  
void soudian()  
{  
    int i,j;  
    for(i=1;i<=scc_cnt;i++)  
        in[i]=out[i]=0;  
    for(i=1;i<=n;i++)  
    {  
        for(j=0;j<g[i].size();j++)  
        {  
            int u=sccon[i];  
            int v=sccon[g[i][j]];  
            if(u!=v)  
            {  
                out[u]++;  
                in[v]++;  
            }  
        }  
    }  
}  
int main()  
{  
    int t,x,y;  
    scanf("%d",&t);  
    while(t--)  
    {  
        scanf("%d%d",&n,&m);  
        for(int i=1;i<=n;i++)  
            g[i].clear();  
        while(m--)  
        {  
            scanf("%d%d",&x,&y);  
            g[x].push_back(y);  
        }  
        memset(low,0,sizeof(low));  
        memset(dfn,0,sizeof(dfn));  
        memset(sccon,0,sizeof(sccon));  
        memset(instack,false,sizeof(instack));  
        dfs_clock=scc_cnt=0;  
        for(int i=1;i<=n;i++)  
            if(!dfn[i])  
                tarjan(i,-1);   
        soudian();  
        if(scc_cnt==1)  
        {  
            printf("0\n");  
            continue;
        }  
        int si=0,so=0;  
        for(int i=1;i<=scc_cnt;i++)  
        {  
            if(in[i]==0)  
                si++;  
            if(out[i]==0)  
                so++;  
        }  
        printf("%d\n",max(si,so));  
    }  
}

问答

19.46%
500ms
131072K

You have two strings

Now an efficient program is required to maintain a operation and support a query.

The operation

The query

Input Format

The first line contains an integer

For each test case, the first line contains an integer $\leq 100000)N (N≤100000).$

The second line is the string $\leq 100000)S (∣S∣≤100000) and the third line is the string T~(|T| \leq 10)T (∣T∣≤10).$

Each of the following

Output Format

For each query, output an integer correponding to the answer.

Output an empty line after each test case.

样例输入

1
5
AABBABA
AA
Q 1 3
C 6 A
Q 2 7
C 2 B
Q 1 5

样例输出

1
2
0

题目来源

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

这个题看起来挺简单的，大概就是区间修改区间查询

直接线段树就可以的，但是有些线段树的常数有点大，可以换个姿势比如树状数组水过的

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[100001];
#define NN 100010

struct node{
       int l, r, cover;
       int sum;
       int key;
}st[NN * 8];

long long f[NN + 2];

void Init(int l, int r, int id){
     st[id].l = l;
     st[id].r = r;
     st[id].cover = 0;
     st[id].key = 0;
     
     if (r - l <= 1){
        st[id].cover = 1;
        st[id].sum = f[l];
        return;
     }
     
     int mid = (l + r) >> 1;
     
     Init(l, mid, id * 2);
     Init(mid, r, id * 2 + 1);
     st[id].sum = st[id * 2].sum + st[id * 2 + 1].sum; //key1
}

void Update(int l, int r, int key, int id){
    
    //key2
     if (st[id].l == l && st[id].r == r){
        st[id].key += key;
        st[id].cover = 1;
        return ;
     }
     st[id].sum += (r - l) * key; //key3
   /*  if (st[id].cover > 0){
        st[id * 2].cover = 1;
        st[id * 2 + 1].cover = 1;
        st[id * 2].key += st[id].key;
        st[id * 2 + 1].key += st[id].key;
        st[id].cover = 0;
        st[id].key = 0;
     }*/
     int mid = (st[id].l + st[id].r) >> 1;
     
     if (r <= mid){
        Update(l, r, key, id * 2);
     }else if(l >= mid){
             Update(l, r, key, id * 2 + 1);
     }else{
           Update(l, mid, key, id * 2);
           Update(mid, r, key, id * 2 + 1);
     }
}

int Search(int l, int r, int id){
        int ans = 0;
        //key4
        if (st[id].cover > 0){
           ans += st[id].key * (r - l);
        }
        
        //key5
        if (st[id].l == l && st[id].r == r){
            ans += st[id].sum;
            return ans;
        }

        int mid = (st[id].l + st[id].r) >> 1;
        
        if (r <= mid){
           return ans + Search(l, r, id * 2);
        }else if(l >= mid){
           return  ans + Search(l, r, id * 2 + 1);
        }else{
           return ans + Search(l, mid, id * 2) + Search(mid, r, id * 2 + 1);
        }
}
int main()
{
    int t,i,j,s,x,r,m,n,y,k;
    char c[5],b[15],e[5];
    scanf("%d",&t);
    while(t--)
    {
        memset(f,0,sizeof(f));
        scanf("%d",&n);
        scanf("%s%s",a,b);
        r=strlen(a);
        m=strlen(b);
        for(i=0;i+m-1<r;i++)
        {
            for(j=0;j<m;j++)
            {
                if(a[i+j]!=b[j])
                break;
            }
            if(j==m)f[i]=1;
        }
        Init(0,r,1);
        //printf("%d\n",s);
        while(n--)
        {
            scanf("%s",c);
            if(c[0]==‘Q‘)
            {
                scanf("%d%d",&x,&y);
                if(y-x+1<m)s=0;
                else s=Search(x-1,y-m+1,1);
                printf("%d\n",s);
            }
            else if(c[0]==‘C‘)
            {
                scanf("%d%s",&x,e);
                a[x-1]=e[0];
                if(x-m<0)i=0;
                else i=x-m;
                for(;i<x&&i+m-1<r;i++)
                {
                    for(j=0;j<m;j++)
                    {
                        if(a[i+j]!=b[j])
                        break;
                    }
                    if(f[i]==1&&j==m||f[i]==0&&j!=m)
                    continue;
                    int u=i+1;
                    if(j!=m)
                    {
                        f[i]=0;
                        Update(u-1,u,-1,1);
                    }
                    if(j==m)
                    {
                        f[i]=1;
                        Update(u-1,u,1,1);
                    }
                }
            }
        }
        puts("");
    }
}

问答

23.59%
1000ms
131072K

In this winter holiday, Bob has a plan for skiing at the mountain resort.

This ski resort has

The $S_iS?i??-th flag to the T_iT?i??-th flag has length L_iL?i??.$

Each path must follow the principal of reduction of heights and the start point must be higher than the end point strictly.

An available ski trail would start from a flag, passing through several flags along the paths, and end at another flag.

Now, you should help Bob find the longest available ski trail in the ski resort.

Input Format

The first line contains an integer

In each test case, the first line contains two integers $\leq 100000<N≤10000and 0 < M \leq 1000000<M≤100000 as described above.$

Each of the following $S_iS?i??, T_iT?i??, and L_i~(0 < L_i < 1000)L?i?? (0<L?i??<1000)describing a path in the ski resort.$

Output Format

For each test case, ouput one integer representing the length of the longest ski trail.

样例输入

样例输出

题目来源

2017 ACM-ICPC 亚洲区（乌鲁木齐赛区）网络赛

最长路，继续dfs维护

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=10003;
struct edge
{
    int to,w;
};
vector<edge>G[N];
ll dis[N];
ll dfs(int u)
{
    if(dis[u])return dis[u];
    ll res=0;
    for(int i=0;i<(int)G[u].size();i++)
    {
        res=max(res,dfs(G[u][i].to)+G[u][i].w);
    }
    return dis[u]=res;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            G[i].clear();
        for(int i=1;i<=m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            G[u].push_back({v,w});
        }
        ll ans=0;
        memset(dis,0,sizeof(ll)*(n+5));
        for(int i=1;i<=n;i++)
        {
            ans=max(dfs(i),ans);
        }
        printf("%lld\n",ans);
    }
    return 0;
}