[hihocoder][Offer收割]编程练习赛45

互补二元组

Xi + Xj = Yi + Yj等价于Xi - Yi + Xj - Yj = 0 ，对每个二元组计算其x与y的差，每次加上其相反数的个数。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

map<int, int> mp;

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(0), cin.tie(0);
    int n, x, y;
    lint ans = 0;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> x >> y;
        if (mp.find(y - x) != mp.end()) ans += mp[y - x];
        mp[x - y]++;
    }
    cout << ans << endl;
    return 0;
}

寻找切线

找到横坐标最大的点，如果这样的点不止一个，直接输出2个。否则计算其他所有点到这一点的斜率，取最大或最小的那个点。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

double k[110000], x[110000], y[110000];

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    std::ios::sync_with_stdio(0), cin.tie(0);
    int n, i1 = 0, i2 = -1;
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> y[i];
        if (x[i] >= x[i1]) i1 = i;
    }
    for (int i = 0; i < n; i++) {
        if (x[i] == x[i1] && i != i1) i2 = i;
    }
    if (i2 != -1) {
        if (i1 > i2) swap(i1, i2);
        cout << i1 + 1 << ' ' << i2 + 1 << endl;
        return 0;
    }
    for (int i = 0; i < n; i++) {
        if (i == i1) k[i] = -1e10;
        else k[i] = (y[i] - y[i1]) / (x[i] - x[i1]);
    }
    i2 = 0;
    for (int i = 0; i < n; i++) {
        if (k[i] > k[i2]) i2 = i;
    }
    if (i1 > i2) swap(i1, i2);
    cout << i1 + 1 << ' ' << i2 + 1 << endl;
    return 0;
}

LR问题

这题出的有问题，因为并不存在什么“最少多少步”的说法，能的话步数是一定的，不能就不能。首先，不看_的话，两个串L和R的顺序应该是一样的。s中L的位置应在t中其对应的L的位置的右边，R在左边，每次加上对应位置坐标的差值就可以。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

char s[110000], t[110000];
VI vs, vt;

int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    //std::ios::sync_with_stdio(0), cin.tie(0);
    scanf("%s", s);
    scanf("%s", t);
    int n = strlen(s);
    vs.clear();
    vt.clear();
    for (int i = 0; i < n; i++) {
        if (s[i] != '_') vs.push_back(s[i]);
        if (t[i] != '_') vt.push_back(t[i]);
    }
    bool ok = true;
    if (vs.size() != vt.size()) ok = false;
    for (int i = 0; i < min(vs.size(), vt.size()); i++) {
        if (vs[i] != vt[i]) {
            ok = false;
            break;
        }
    }
    if (!ok) {
        cout << -1 << endl;
        return 0;
    }
    int ans = 0;
    vs.clear();
    vt.clear();
    for (int i = 0; i < n; i++) {
        if (s[i] == 'L') vs.push_back(i);
        if (t[i] == 'L') vt.push_back(i);
    }
    for (int i = 0; i < min(vs.size(), vt.size()); i++) {
        if (vs[i] < vt[i]) ok = false;
        else ans += vs[i] - vt[i];
    }
    vs.clear();
    vt.clear();
    for (int i = 0; i < n; i++) {
        if (s[i] == 'R') vs.push_back(i);
        if (t[i] == 'R') vt.push_back(i);
    }
    for (int i = 0; i < min(vs.size(), vt.size()); i++) {
        if (vs[i] > vt[i]) ok = false;
        else ans += vt[i] - vs[i];
    }
    if (!ok) cout << -1 << endl;
    else cout << ans << endl;
    return 0;
}

推断大小关系

对于x<y，连一条从x指向y的边；对于x=y，连一条从x指向y的边和从y指向x的边。当存在从1到n或从n到1的路径时，能推断大小关系。

#pragma comment(linker, "/STACK:102400000,102400000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<algorithm>
#include<iostream>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<functional>
#include<math.h>
//#include<bits/stdc++.h>
using namespace std;
typedef long long lint;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef queue<int> QI;


void makedata() {
    freopen("input.txt", "w", stdout);
    fclose(stdout);
}

bool f1[110000], fn[110000];
VI G[110000];
void dfs1(int x) {
    if (f1[x]) return;
    f1[x] = true;
    for (int i = 0; i < G[x].size(); i++) dfs1(G[x][i]);
}

void dfsn(int x) {
    if (fn[x]) return;
    fn[x] = true;
    for (int i = 0; i < G[x].size(); i++) dfsn(G[x][i]);
}


int main() {
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
#endif
    //makedata();
    //std::ios::sync_with_stdio(0), cin.tie(0);
    int n, m, x, y;
    char ch;
    memset(f1, false, sizeof(f1));
    memset(fn, false, sizeof(fn));
    scanf("%d%d
", &n, &m);
    f1[1] = true, fn[n] = true;
    for (int i = 1; i <= m; i++) {
        scanf("A%d %c A%d
", &x, &ch, &y);
        G[x].push_back(y);
        if (f1[x]) dfs1(y);
        if (fn[x]) dfsn(y);
        if (ch == '=') {
            G[y].push_back(x);
            if (f1[y]) dfs1(x);
            if (fn[y]) dfsn(x);
        }
        if (f1[n] || fn[1]) {
            printf("%d
", i);
            return 0;
        }
    }
    printf("-1
");
    return 0;
}