2013-2014 ACM-ICPC, NEERC, Eastern Subregional Contest PART (7/10)

Posted 2021-03-13 kikokiko

[2013-2014 ACM-ICPC, NEERC, Eastern Subregional Contest]

(A.Podracing)

(B.The battle near the swamp)

签到

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e4+7;
int n,m;
int main(){
    scanf("%d %d",&n,&m);
    int lft = 0, ene = 0;
    for(int x,i = 1; i <= n; i++){
        scanf("%d",&x);
        if(x>m) lft+=x-m;
        else ene+=m-x;
    }
    cout << lft << ' ' << ene << endl;
    return 0;
}

(C.CVS)

可持久化链表，可以用链式前向星

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 5e5+7;
struct CVS{
    int tot;
    CVS(){ tot = 1; }
    struct NODE{
        NODE *pre;
        int ID;
        NODE(int id = 0){
            pre = nullptr;
            ID = id;
        }
    };
    struct Kamino{
        NODE *ltail,*ttail;
        Kamino(){
            ltail = new NODE;
            ttail = new NODE;
        }
    }kam[MAXN];
    void learn(int ID, int p){
        kam[ID].ttail = new NODE;
        NODE* node = new NODE(p);
        node->pre = kam[ID].ltail;
        kam[ID].ltail = node;
    }
    void rollback(int ID){
        NODE* node = new NODE(kam[ID].ltail->ID);
        node->pre = kam[ID].ttail;
        kam[ID].ttail = node;
        kam[ID].ltail = kam[ID].ltail->pre;
    }
    void relearn(int ID){
        NODE* node = new NODE(kam[ID].ttail->ID);
        node->pre = kam[ID].ltail;
        kam[ID].ltail = node;
        kam[ID].ttail = kam[ID].ttail->pre;
    }
    void clone(int ID){
        tot++;
        kam[tot] = kam[ID];
    }
    int check(int ID){
        return kam[ID].ltail->ID;
    }
}CVS;
int n,m;
int read(){
    int x = 0, f = 1;
    char c = getchar();
    while(c!='-'&&(c<'0'||c>'9')) c = getchar();
    if(c=='-') f = -1,c = getchar();
    while(c>='0'&&c<='9') x = x*10+c-'0', c = getchar();
    return f*x;
}
int main(){
    n = read(); m = read();
    while(n--){
        char op[20];
        scanf("%s",op);
        if(op[0]=='l'){         //learn
            int id=read(),p=read();
            CVS.learn(id,p);
        }
        else if(op[1]=='o'){    //rollback
            int id=read();
            CVS.rollback(id);
        }
        else if(op[1]=='e'){    //relearn
            int id=read();
            CVS.relearn(id);
        }
        else if(op[1]=='l'){    //clone
            int id=read();
            CVS.clone(id);
        }
        else if(op[1]=='h'){    //check
            int id=read();
            if(CVS.check(id)) printf("%d
",CVS.check(id));
            else printf("basic
");
        }
    }
    return 0;
}

(D.This cheeseburger you don't need)

模拟

#include<bits/stdc++.h>
using namespace std;
#define IOS std::ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
const int inf = 0x3f3f3f3f;
string S[5];
int main(){
    IOS;
    string s;
    string ans="";
    int tar=0;
    for(int i=0;i<3;i++)S[i]="";
    int f=0;
    while(cin>>s){
        int len=s.length();
        if(s[0]=='(')tar=0,f++;
        if(s[0]=='[')tar=1,f++;
        if(s[0]=='{')tar=2,f++;
        if(s[len-1]==')')f++;
        if(s[len-1]==']')f++;
        if(s[len-1]=='}')f++;
        if(s[len-2]==']')f++;
        if(s[len-2]=='}')f++;
        if(s[len-2]==')')f++;
        if(f==0){
            ans+=s+' ';
            continue;
        }
        if(s[0]!='('&&s[0]!='['&&s[0]!='{')S[tar]+=s[0];
        if(s[len-1]==',')S[tar]+=s.substr(1,max(0,len-3));
        else S[tar]+=s.substr(1,max(0,len-2));
        if(s[len-1]==','){
            if(len>=3&&s[len-2]!=')'&&s[len-2]!=']'&&s[len-2]!='}')S[tar]+=s[len-2];
            S[tar]+=' ';
            if(f==6){
                ans+=S[2];
                ans+=S[0];
                ans+=S[1].substr(0,S[1].length()-1)+", ";
                f=0;
                for(int i=0;i<3;i++)S[i].clear();
            }
            continue;
        }
        if(s[len-1]!=')'&&s[len-1]!=']'&&s[len-1]!='}'&&len!=1)S[tar]+=s[len-1];
        S[tar]+=' ';
        if(f==6){
            ans+=S[2];
            ans+=S[0];
            ans+=S[1];
            f=0;
            for(int i=0;i<3;i++)S[i].clear();
        }
    }
    int len=ans.length();
    for(int i=0;i<len-1;i++){
        if(i==0&&ans[i]>='a'&&ans[i]<='z')cout<<char(ans[i]-32);
        else if(i!=0 && ans[i]>='A'&&ans[i]<='Z')cout<<char(ans[i]+32);
        else cout<<ans[i];
    }
    return 0;
}

(E.The Emperor's plan)

组合数学 概率DP
(f[x][y])表示当天晚上参议员中非spy的有(x)个，spy有(y)个的情况下，最终剩余非spy参议员的期望数量
边界条件是
1.(x le y)这时非spy在晚上全被消灭，(f[x][y]=0)
2.(y==0) 这时全为非spy参议员，(f[x][y]=x)
然后依据题意进行(dp)，每次枚举白天被票出的人的个数进行转移，用(ln)和(exp)保证精度正确

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 222;
double ln[MAXN],f[MAXN][MAXN];
double lnC(int A, int B){ return ln[A]-ln[B]-ln[A-B]; }
double percentage(int a, int x, int b, int y){ return exp(lnC(x,a)+lnC(y,b)-lnC(x+y,a+b)); }
double solve(int x, int y){
    if(f[x][y]!=-1) return f[x][y];
    if(x<=y) return f[x][y] = 0;
    if(y==0) return f[x][y] = x;
    x-=y;
    int tot = x+y;
    double res = 0;
    for(int k = 1; k < tot; k++){
        double tp = 0;
        for(int i = max(0,k-y); i <= min(x,k); i++) tp += solve(x-i,y-k+i)*percentage(i,x,k-i,y);
        res = max(res,tp);
    }
    return f[x+y][y] = res;
}
int n,k;
int main(){
    ln[0] = 0;
    for(int i = 1; i < MAXN; i++) ln[i] = ln[i-1]+log(i);
    for(int  i = 0; i < MAXN; i++) for(int j = 0; j < MAXN; j++) f[i][j] = -1;
    scanf("%d %d",&n,&k);
    printf("%.6f
",solve(n-k,k));
    return 0;
}

(F.Illegal spices)

题目保证答案存在，那就把前(n-m)个都设为(1)，后面的(m)个贪心地取就好了

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
int n,m,p;
vector<int> vec;
int main(){
    scanf("%d %d %d",&n,&m,&p);
    int_fast64_t sum = n - m;
    int num = n - m, cur = 2;
    for(int i = 1; i <= n-m; i++) vec.emplace_back(1);
    for(int i = n-m+1; i <= n; i++){
        if(num*100<p*(i-1)){
            cur++;
            num = i - 1;
        }
        sum+=cur;
        vec.emplace_back(cur);    
    }
    cout << sum << endl;
    for(int x : vec) cout << x << ' ';
    return 0;
}

(G.Cipher Message 3)

(H.Those are not the droids you're looking for)

出入的时间分开跑二分图匹配即可，如果能完全匹配则说明没有说谎

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1111;
int a,b,n,match[MAXN];
bool vis[MAXN];
vector<int> G[MAXN];
pair<int,int> sta[MAXN];
bool dfs(int u){
    vis[u] = true;
    for(int v : G[u]){
        if(match[v]==-1||(!vis[match[v]]&&dfs(match[v]))){
            match[v] = u;
            return true;
        }
    }
    return false;
}
int go(){
    memset(match,255,sizeof(match));
    int tot = 0;
    for(int i = 1; i <= n; i++){
        if(sta[i].second==0){
            memset(vis,0,sizeof(vis));
            if(dfs(i)) tot++;
            else return tot;
        }
    }
    return tot;
}
int main(){
    scanf("%d %d %d",&a,&b,&n);
    for(int i = 1; i <= n; i++) scanf("%d %d",&sta[i].first,&sta[i].second);
    for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++){
        if(sta[i].second==0&&sta[j].second==1&&(sta[j].first-sta[i].first>=a||(sta[j].first-sta[i].first<=b&&sta[j].first-sta[i].first>0))){
            G[i].emplace_back(j);
        }
    }
    if(go()*2==n){
        printf("No reason
");
        for(int i = 1; i <= n; i++) if(sta[i].second==1){
            printf("%d %d
",sta[match[i]].first,sta[i].first);
        }
    }
    else printf("Liar
");
    return 0;
}

(I.The old Padawan)

先双指针将每个点会回到的点找出来，然后暴力就行了

//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
const int MAXN = 1e5+7;
int n,m,cur;
LL k,A[MAXN],tk[MAXN],res,pos[MAXN];
int main(){
    scanf("%d %d %I64d",&n,&m,&k);
    for(int i = 1; i <= n; i++) scanf("%I64d",&A[i]);
    int l = 0,r = 1;
    LL sum = A[1];
    while(r<=n){
        while(sum-A[l+1]>k) sum-=A[++l];
        pos[r] = l;
        sum+=A[++r];
    }
    for(int i = 1; i <= m; i++) scanf("%I64d",&tk[i]);
    for(int i = 1; i <= m; i++){
        if(n-cur<tk[i]-res) break;
        int dur = tk[i] - res - 1;
        res += dur+1;
        cur += dur;
        cur = pos[cur];
    }
    if(cur<n) res+=n-cur;
    cout << res << endl;
    return 0;
}

(J.The secret module)