BUPT2017 springtraining(16) #6 ——图论

题目链接

A.容易发现最后字符的对应都是一对一的

或者说我们没办法出现最后多对一或者一对多的情况

所以只要算出 ‘a’ - 'z' 每个字符最后对应的字符即可

#include <cstdio>
#include <algorithm>

int n, m, b[26];

char a[26], s[200010];

int main() {
    char s1[5], s2[5];
    scanf("%d %d %s", &n, &m, s);
    for(int i = 0;i < 26;i ++) a[i] = i + 'a', b[i] = i;
    while(m --) {
        scanf("%s %s", s1, s2);
        if(s1[0] == s2[0]) continue;
        a[b[s1[0] - 'a']] = s2[0];
        a[b[s2[0] - 'a']] = s1[0];
        std::swap(b[s1[0] - 'a'], b[s2[0] - 'a']);
    }
    for(int i = 0;i < n;i ++)
        putchar(a[s[i] - 'a']);
    return 0;
}

B.

C.A simple directed graph is a directed graph having no multiple edges or graph loops.

graph loops其实说的是自环?道理我都懂，但是自环不是self loops之类的吗？

我们考虑连完所有边的最终形态，应该是

两个强连通分量AB，内部均为完全图

A中所有点都能到B中所有点

B中没有连向A中点的边

稍微算一下可知令siz_A与siz_B差距越大，可连边就越多

那我们先对原图求强连通分量

这里面没有入度或没有出度的scc是有可能成为A或B的scc

找出这里面siz最小的scc来确定A和B其中一个的siz即可

为什么不可能是siz最大的一个呢？

我们列一下式子发现时恒等式

siz最小的必然不比siz最大的结果差

#include <bits/stdc++.h>

using namespace std;

const int maxn = 100010;

int n, m, cnt, top, sum, c1[maxn], c2[maxn], siz[maxn];

int in[maxn], vis[maxn], sta[maxn], dfn[maxn], low[maxn], bel[maxn];

vector <int> e[maxn];

void tarjan(int x) {
    vis[x] = 1;
    in[x] = 1, sta[++ top] = x, dfn[x] = low[x] = ++ cnt;
    for(int t, i = 0;i < e[x].size();i ++) {
        t = e[x][i];
        if(!vis[t]) {
            tarjan(t);
            low[x] = min(low[x], low[t]);
        }
        else if(in[t] && low[t] < low[x]) low[x] = low[t];
    }
    if(low[x] == dfn[x]) {
        sum ++;
        for(int i = 0;i != x;i = sta[top --], bel[i] = sum, in[i] = 0, siz[sum] ++);
    }
}

int main() {
    long long ans;
    int Case, u, v, tmp;
    ios::sync_with_stdio(false);
    cin >> Case;
    for(int tt = 1;tt <= Case;tt ++) {
        cout << "Case " << tt << ": ";
        cin >> n >> m;
        cnt = ans = sum = 0, tmp = n;
        for(int i = 1;i <= n;i ++) e[i].clear(), vis[i] = c1[i] = c2[i] = siz[i] = 0;
        for(int i = 1;i <= m;i ++) cin >> u >> v, e[u].push_back(v);
        for(int i = 1;i <= n;i ++)
            if(!vis[i]) tarjan(i);
        if(sum == 1) cout << -1 << endl;
        else {
            for(int i = 1;i <= n;i ++)
                for(int j = 0;j < e[i].size();j ++) 
                    if(bel[i] != bel[e[i][j]]) c1[bel[i]] ++, c2[bel[e[i][j]]] ++;
            for(int i = 1;i <= sum;i ++) 
                if(c1[i] == 0 || c2[i] == 0) 
                    tmp = min(tmp, siz[i]);
             cout << 1ll * n * n - m - n - 1ll * tmp * (n - tmp) << endl;
        }
    }
}

D.其实只要先处理出昨天没走的人

就可以算出今天各个时刻的人数了，取max即可

#include <iostream>

using namespace std;

bool in[1000010];

int n, ans, cnt, x;

char str[110][3];

int main() {
    ios::sync_with_stdio(false);
    cin >> n;
    for(int i = 1;i <= n;i ++) {
        cin >> str[i] >> x;
        if(str[i][0] == '+') in[x] = 1;
        else if(!in[x]) cnt ++;
    }
    ans = cnt;
    for(int i = 1;i <= n;i ++) {
        if(str[i][0] == '+') cnt ++, ans = max(cnt, ans);
        else cnt --;
    }
    cout << ans;
}

E.我们令第 i 个数为 a[i]

计算出以 a[i] 为序列第一个数的方案种数

以 a[i] 为序列第二个数的方案种数

ans += 以 a[i] / k 为序列第二个数的方案种数

#include <map>
#include <iostream>

using namespace std;

typedef long long ll;

map <ll, ll> p[2];

int n, k;

int a[200010];

ll ans;

int main() {
    ios::sync_with_stdio(false);
    cin >> n >> k;
    for(int i = 1;i <= n;i ++) {
        cin >> a[i];
        if(a[i] % k == 0) ans += p[1][a[i] / k], p[1][a[i]] += p[0][a[i] / k];
        p[0][a[i]] ++;
    }
    cout << ans;
    return 0;
}

F.

G.先缩点

#include <queue>
#include <vector>
#include <cstring>
#include <iostream>

using namespace std;

const int maxn = 100010;

struct node{
    int x, y;
    bool operator < (const node &a) const {
        return y > a.y;
    }
};

vector <node> e[maxn], f[maxn];

priority_queue <node> q;

int n, N, m, vi[maxn];

long long ans;

int cnt, top, in[maxn], dfn[maxn], low[maxn], sta[maxn], vis[maxn], bel[maxn], sum[maxn];

void Clear() {
    ans = cnt = N = 0;
    for(int i = 0;i < n;i ++) e[i].clear(), f[i].clear(), vi[i] = vis[i] = 0;
    while(!q.empty()) q.pop();
}

void tarjan(int x, int t = 0) {
    vis[x] = 1;
    dfn[x] = low[x] = ++ cnt, sta[++ top] = x, in[x] = 1;
    for(int i = 0;i < e[x].size();i ++) {
        t = e[x][i].x;
        if(!vis[t]) {
            tarjan(t);
            low[x] = min(low[x], low[t]);
        }
        else if(in[t] && low[t] < low[x]) low[x] = low[t]; 
    }
    if(low[x] == dfn[x]) {
        for(int i = -1;i != x;i = sta[top --], bel[i] = N, in[i] = 0);
        N ++;
    }
}

int main() {
    ios::sync_with_stdio(false);
    int u, v, w;
    node tmp;
    while(cin >> n >> m) {
        Clear();
        for(int i = 1;i <= m;i ++)
            cin >> u >> v >> w, e[u].push_back((node){v, w});
        for(int i = 0;i < n;i ++)
            if(!vis[i]) tarjan(i);
        memset(sum, 0x3f,sizeof sum);
        for(int i = 0;i < n;i ++) {
            for(int j = 0;j < e[i].size();j ++)
                if(bel[i] != bel[e[i][j].x])
                    sum[bel[e[i][j].x]] = min(sum[bel[e[i][j].x]], e[i][j].y);
        }
        for(int i = 0;i < N;i ++)
            if(sum[i] != 0x3f3f3f3f)
                ans += sum[i];
        cout << ans << endl;
    } 
    return 0;
}

H.手写一下发现总共只有2 + 4 + 8 + ... + 2^9 = 1022 个数

所以枚举对比一下就好了

#include <vector>
#include <iostream>

using namespace std;

vector <int> e[10];

int n, sum;

int main() {
    e[0].push_back(0);
    cin >> n;
    for(int i = 1, j = 1;i <= 9;i ++, j *= 10) {
        for(int t, k = 0;k < e[i - 1].size();k ++) {
            t = 4 * j + e[i - 1][k];
            e[i].push_back(t);
            if(t == n) {
                cout << sum + e[i].size();
                return 0;
            }
        }
        for(int t, k = 0;k < e[i - 1].size();k ++) {
            t = 7 * j + e[i - 1][k];
            e[i].push_back(t);
            if(t == n) {
                cout << sum + e[i].size();
                return 0;
            }
        }
        sum += e[i].size();
    }
}