Codeforces Beta Round #19

概述  A. World Football Cup #include <bits/stdc++.h>using namespace std; const int N = 60;char name[N][N];map<string, int> mp;char s[N]; struct P { int id, point; int dif, goal;

 

A. World Football Cup

#include <bits/stdc++.h>using namespace std; const int N = 60;char name[N][N];map<string,int> mp;char s[N]; struct P {    int ID,point;    int dif,goal;    bool operator < (const P &rhs) const {        if (point == rhs.point && goal - dif == rhs.goal - rhs.dif) return goal > rhs.goal;        if (point == rhs.point) return goal - dif > rhs.goal - rhs.dif;        return point > rhs.point;    }} p[N]; int main() {    int n;    scanf("%d",&n);    for (int i = 0; i < n; i++) {        scanf("%s",name[i]);        mp[name[i]] = i;        p[i].ID = i;    }    for (int i = 0; i < n * (n - 1) / 2; i++) {        int pp,qq;        scanf("%s%d:%d",s,&pp,&qq);        int j = 0;        int len = strlen(s);        char s1[50] = {};        char s2[50] = {};        int p1 = 0,p2 = 0;        for (; s[j] != ‘-‘; j++) s1[p1++] = s[j];        j++;        for (; j < len; j++) s2[p2++] = s[j];        int a = mp[s1],b = mp[s2];        p[a].goal += pp,p[b].goal += qq;        p[a].dif += qq; p[b].dif += pp;        if (pp > qq) p[a].point += 3;        else if (pp == qq) p[a].point++,p[b].point++;        else p[b].point += 3;    }    sort(p,p + n);    vector<string> ans;    for (int i = 0; i < n / 2; i++)        ans.push_back(name[p[i].ID]);    sort(ans.begin(),ans.end());    for (auto x: ans)        cout << x << ‘\n‘;    return 0;}

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B. Checkout Assistant

有$n$个商品，每个商品价格$c_i$，售货员处理这件商品需要时间$t_i$，偷走一个商品需要$1$个单位时间，问怎么安排这些商品的结账顺序可以使的最后要还的钱最少。

如果让一件商品去结账，相当于花$c_i$块钱，能偷走买走至多$t_i + 1$个商品，要使最后总花费最小，那么就是01背包了。

#include <bits/stdc++.h>#define ll long longusing namespace std; const int N = 2020;ll dp[N];int w[N];ll c[N]; int main() {    memset(dp,0x3f,sizeof dp);    dp[0] = 0;    int n;    scanf("%d",&n);    for (int i = 1; i <= n; i++)        scanf("%d%lld",&w[i],&c[i]),w[i]++;    for (int i = 1; i <= n; i++) {        for (int j = n; j >= 1; j--) {            dp[j] = min(dp[j],dp[max(j - w[i],0)] + c[i]);        }    }    printf("%lld\n",dp[n]);    return 0;}

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C. Deletion of Repeats

有一个序列，每种元素至多出现$10$次，如果有一个子串前一半等于后一半，那么把前一半及以前的字符全部删掉，要求从短的以及较左的地方开始删，问最后这个串长什么样。

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 7;int a[N],v[N];vector<int> G[N];int main() {    int n;    scanf("%d",&n);    for (int i = 1; i <= n; i++) {        scanf("%d",&a[i]);        v[i] = a[i];    }    sort(v + 1,v + 1 + n);    int cnt = unique(v + 1,v + 1 + n) - v - 1;    for (int i = 1; i <= n; i++) {        a[i] = lower_bound(v + 1,v + 1 + cnt,a[i]) - v;        G[a[i]].push_back(i);    }    int Now = 0;    for (int i = 2; i <= n; i++) {        bool flag = 0;        for (auto pos: G[a[i]]) {            flag = 1;            if (pos >= i) continue;            if (n - i + 1 < i - pos) continue;            if (pos <= Now) continue;            flag = 0;            for (int j = 1; j < i - pos; j++) {                if (a[i + j] != a[pos + j]) {                    flag = 1;                    break;                }            }            if (!flag) break;        }        if (!flag) Now = i - 1;    }    printf("%d\n",n - Now);    for (int i = Now + 1; i <= n; i++)        printf("%d%c",v[a[i]]," \n"[i == n]);    return 0;}

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D. Points

一个二维坐标系，有三个 *** 作。

1.加点

2.删点

3.查询严格在$(x,y)$右上角的点，输出$x$坐标最小的那个点，如果有相同的，则输出$y$最小的，不存在输出$-1$

因为存在修改 *** 作，那么就不能离线做了，用线段树维护$x$坐标下$y$坐标的最大值，然后就变成了查询$\left[x + 1,inf\right]$中最小的$x$其$y$大于要查询的$y$，那么就又是先查左及剪枝的那个写法了。然后找到对应$x$坐标后，加点删点用一个set维护每个$x$坐标出现过$y$的坐标，在这个set里面lower_bound一下就好了，刚开始都插入0以及所有$y$坐标都加个一会好 *** 作很多。

#include <bits/stdc++.h>using namespace std; const int N = 5e5 + 7; struct Seg {    #define lp p << 1    #define rp p << 1 | 1    int mx[N << 2];    voID pushup(int p) {        mx[p] = max(mx[lp],mx[rp]);    }    voID update(int p,int l,int r,int x,int val) {        if (l == r) {            mx[p] = max(val,mx[p]);            return;        }        int mID = l + r >> 1;        if (x <= mID) update(lp,l,mID,x,val);        else update(rp,mID + 1,r,val);        pushup(p);    }    voID change(int p,int val) {        if (l == r) {            mx[p] = val;            return;        }        int mID = l + r >> 1;        if (x <= mID) change(lp,val);        else change(rp,val);        pushup(p);    }    int query(int p,int y,int val) {        if (x > y) return -1;        if (mx[p] <= val) return -1;        if (l == r) return l;        int mID = l + r >> 1;        if (x <= mID) {            int temp = query(lp,y,val);            if (temp != -1) return temp;        }        return query(rp,val);    }} seg; struct OPT {    int opt;    int x,y;} o[N];int v[N];set<int> st[N]; int main() {//    freopen("in.txt","r",stdin);    int n;    scanf("%d",&n);    for (int i = 1; i <= n; i++) {        char s[10] = {};        scanf("%s%d%d",&o[i].x,&o[i].y);        if (s[0] == ‘a‘) o[i].opt = 0;        else if (s[0] == ‘r‘) o[i].opt = 1;        else o[i].opt = 2;        v[i] = o[i].x;        o[i].y++;    }    sort(v + 1,v + 1 + n) - v - 1;    for (int i = 1; i <= cnt; i++) st[i].insert(0);    for (int i = 1; i <= n; i++)         o[i].x = lower_bound(v + 1,o[i].x) - v;    for (int i = 1; i <= n; i++) {        if (o[i].opt == 0) {            st[o[i].x].insert(o[i].y);            seg.update(1,1,cnt,o[i].x,o[i].y);        } else if (o[i].opt == 1) {            st[o[i].x].erase(o[i].y);            set<int>::iterator it = st[o[i].x].end();            it--;            int res = 0;            res = *it;            seg.change(1,res);        } else {            int pos = seg.query(1,1,o[i].x + 1,o[i].y);            if (pos == -1) printf("-1\n");            else {                int ans = *st[pos].upper_bound(o[i].y);                printf("%d %d\n",v[pos],ans - 1);            }        }    }    return 0;}

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