Codeforces Round #375 (Div. 2) ABCDE

A - The New Year: Meeting Friends

水

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int a,b,c;
    cin>>a>>b>>c;
    cout<<max(a,max(b,c)) - min(a,min(b,c)) <<endl;
    return 0;
}

B - Text Document Analysis

字符串暴力模拟，感觉还是需要一点技巧的，我写的太慢了。

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <iostream>

using namespace std;
const int N = 260;
char a[N];
//37
//_Hello_Vasya(and_Petya)__bye_(and_OK)

bool is_s(char ch) {
    if (ch == ‘_‘ || ch == ‘(‘ || ch == ‘)‘ || ch == 0) return true;
    return false;
}

int read(char str[]) {
    int idx = 0;
    int ans = 0;
    while (!is_s(str[idx++])) ans++;
    return ans;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int ans1 = 0, ans2 = 0, n = 12;
    scanf("%d", &n);
    scanf("%s", a);
    bool fg = false;
    for (int i = 0; i <= n;) {
        if (a[i] == ‘(‘) fg = true;
        if (a[i] == ‘)‘) fg = false;
        if (is_s(a[i])) {
            i++;
        } else {
            int tmp = read(a+i);
            if (!fg) ans1 = max(ans1, tmp);
            else ans2++;
            i += tmp;
        }
    }
    cout << ans1 << " " << ans2;
    return 0;
}

C - Polycarp at the Radio

应该也是sb暴力题，map乱搞的(队友写的==

#include <cstdio>
#include <map>
#include <queue>
using namespace std;
const int maxn = 2010;
map<int, int> mp;
queue<int> q;
int a[maxn], n, m;
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), mp[a[i]]++;
    int ans = n / m;
    for (int i = 1; i <= m; i++) if (mp[i] < ans) q.push(i);
    int cnt = 0;
    while (!q.empty()) {
        int v = q.front(); q.pop();
        for (int i = 1; i <= n; i++) {
            if (a[i] > m || mp[a[i]] > ans) mp[a[i]]--, a[i] = v, mp[v]++, cnt++;
            if (mp[v] >= ans) break;
        }
    }
    printf("%d %d\n", ans, cnt);
    for (int i = 1; i <= n; i++) printf("%d%c", a[i], i==n?‘\n‘:‘ ‘);
    return 0;
}

D - Lakes in Berland

简单搜索。因为数据小，随便暴力。

#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <queue>
#include <vector>

using namespace std;
const int N = 60;

char mp[N][N];
int vis[N][N], n, m, k;
int dir[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};

int dfs(int x, int y, int cnt) {
    if (x < 0 || x >= n || y < 0 || y >= m) return -1;
    if (mp[x][y] == ‘*‘ || vis[x][y]) return 0;
    vis[x][y] = cnt;
    int ans = 1;
    for (int i = 0; i < 4; ++i) {
        int tmp = dfs(x+dir[i][0], y+dir[i][1], cnt);
        if (ans != -1) {
            if (tmp == -1) ans = -1;
            else ans += tmp;
        }
    }
    return ans;
}

vector<pair<int, int> > g;

int main() {
    while (~scanf("%d%d%d", &n, &m, &k)) {
        for (int i = 0; i < n; ++i) scanf("%s", mp[i]);
        int cnt = 0, tmp, ans = 0;
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                if (!vis[i][j] && mp[i][j] == ‘.‘)
                    if ((tmp = dfs(i, j, ++cnt)) != -1) g.push_back(make_pair(tmp, cnt));
        sort(g.begin(), g.end());
        tmp = g.size() - k;
        for (int i = 0; i < tmp; ++i) ans += g[i].first;
        printf("%d\n", ans);
        for (int k = 0; k < tmp; ++k)
            for (int i = 0; i < n; ++i)
                for (int j = 0; j < m; ++j)
                    if (vis[i][j] == g[k].second) mp[i][j] = ‘*‘;
        for (int i = 0; i < n; ++i) printf("%s\n", mp[i]);
    }
    return 0;
}

E. One-Way Reform

给一个无向图，要求把每个边都变成有向边，使尽可能多的点入度等于出度，问最多能满足多少个点，并输出每条边。（任意顺序）

感觉很厉害的一道题。

无向图存在欧拉回路的充要条件：一个无向图存在欧拉回路，当且仅当该图所有顶点度数都为偶数,且该图是连通图。

有向图存在欧拉回路的充要条件：一个有向图存在欧拉回路，所有顶点的入度等于出度且该图是连通图。

首先可以知道奇数度的点是不可能到达要求的。找出奇数度的点，两两一对连接，记做虚边，这样就可以保证每个点的度数都是偶数，那么就可以找到一条欧拉回路，保证每个点入度等于出度。

因为总度数是偶数，所以奇数度的点一定是偶数个。

如果图不连通，对每一个连通图求欧拉回路就好了。

然后输出图中非虚边就ok了。

好难想到啊= =队友花了半天给我讲明白的……然后还是WA了好多好多次（我是有多蠢啊T^T

#include <bits/stdc++.h>
using namespace std;

const int N = 250;

struct Edge {
    int next, from, to, fg; // fg: 0,2未处理 1选 -1不选
} e[N*N];
int head[N], cntE;
void addedge(int u, int v, int fg) {
    e[cntE].to = v; e[cntE].from = u;
    e[cntE].next = head[u]; e[cntE].fg = fg;
    head[u] = cntE++;
}
int deg[N];

void dfs(int u) {
    for (int i = head[u]; ~i; i = e[i].next) {
        if (e[i].fg == 0) {
            e[i].fg = 1;
            e[i^1].fg = -1;
            dfs(e[i].to);
        } else if (e[i].fg == 2) {
            e[i].fg = e[i^1].fg = -1;
            dfs(e[i].to);
        }
    }
}

int main() {
    #ifndef ONLINE_JUDGE
       freopen("in.txt", "r", stdin);
    #endif // ONLINE_JUDGE

    int T, n, m, u, v;
    scanf("%d", &T);
    while (T--) {
        memset(head, -1, sizeof head); cntE = 0;
        memset(deg, 0, sizeof deg);
        scanf("%d%d", &n, &m);
        for (int i = 0; i < m; ++i) {
            scanf("%d%d", &u, &v);
            addedge(u, v, 0);
            addedge(v, u, 0);
            deg[u]++, deg[v]++;
        }
        int last = 0, odd = 0;
        for (int i = 1; i <= n; ++i) {
            if (deg[i] & 1) {
                ++odd;
                if (last) addedge(last, i, 2), addedge(i, last, 2), last = 0;
                else last = i;
            }
        }
        for (int i = 1; i <= n; ++i) dfs(i);
        printf("%d\n", n - odd);
        for (int i = 0; i < cntE; ++i)
            if (e[i].fg == 1) printf("%d %d\n", e[i].from, e[i].to);
    }
    return 0;
}

F. st-Spanning Tree

挖坑待填……