CodeForces#286 div.2 题解

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本文提供CodeForces 505比赛A-D题的详细解答，包括回文字符串构造、多颜色路径计数、动态规划最大值求解及强连通分量的边数计算。

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题目链接：http://codeforces.com/contest/505/problems

分类：A、枚举；B、BFS；C、DP；D、强联通；E、不明

A题：

题意：在字符串的任意任意位置插入一个字符，使得得到的字符串回文

思路：数据范围很小，字符串长度才10，一共276种情况，枚举就行了。

下面是A题代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);

inline bool judge ( const string str )
{
    int a = 0;
    int b = str.size() - 1;
    for ( ; a <= b ; ){
        if ( str[a] != str[b] ){
            return 0;
        }
        a ++;
        b --;
    }
    return 1;
}

int main()
{
    ios::sync_with_stdio( false );
    string a;
    string b;
    cin >> a;
    int cnt = 0;
    for ( int i = 0; i <= a.size(); i ++ ){
        for ( int j = 0; j < 26; j ++ ){
            b = a;
            string k = "a";
            k[0] = k[0] + j;
            b.insert ( i, k );
            if ( judge ( b ) ){
                cout << b << endl;
                return 0;
            }
        }
    }
    cout << "NA" << endl;
    return 0;
}

B题：

题意：给出一张无向图，每条边都有对应的颜色，并且每次只能利用同一种颜色的边。求从a点移动到b点有多少种方案。

思路：同上题，数据范围相当小，只有100个节点，颜色也不超过100种，直接对a点的进行搜索，枚举有多少种颜色的通路能到b点

下面是B题代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);

const int maxn = 1005;
vector<int> g[maxn][maxn];

bool SPFA ( int be, int en, int k )
{
    queue<int> que;
    que.push ( be );
    int vis[1000] = { 0 };
    while ( ! que.empty() ){
        int tmp = que.front();
        que.pop();
        for ( int i = 0; i < g[k][tmp].size() && vis[tmp] == 0; i ++ ){
            que.push( g[k][tmp][i] );
            if ( g[k][tmp][i] == en )
                return 1;
        }
        vis[tmp] = 1;
    }
    return 0;
}

int main()
{
    ios::sync_with_stdio( false );
    int m, n;
    while ( cin >> m >> n ){
        int a, b, c;
        for ( int i = 0; i < n; i ++ ){
            cin >> a >> b >> c;
            g[c][a].push_back ( b );
            g[c][b].push_back ( a );
        }
        int k;
        cin >> k;
        int be, en;
        for ( int i = 0; i < k; i ++ ){
            cin >> be >> en;
            int cnt = 0;
            for ( int j = 1; j <= n; j ++ ){
                if ( SPFA ( be, en, j ) ) {
                    cnt ++;
                }
            }
            cout << cnt << endl;
        }
    }
    return 0;
}

C题：

题意：从起点移动到终点，每次只能比前一次移动距离+1，-1，或者不变。每移动到一个点就将该点的值累加，求最大的情况。

思路：动态规划，有第一维表示位置，第二维表示步长，并且最长步长不能超过250，（1+2+3+.....+250) > 30000 , 所以第二维第长度为500，状态转移方程为 dp[i][j] = dp[i][j] + max ( dp[i-(j-301+n)][j-1], dp[i-(j-301+n)][j], dp[i-(j-301+n)][j+1] ); ans = max ( ans, f[i][j] );

下面是C题代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);

int num[30005];
int dp[30005][605];
int main()
{
    memset ( dp, -1, sizeof ( dp ) );
    memset ( num, 0, sizeof ( num ) );
    int m, n;
    int k;
    cin >> m >> n;
    for ( int i = 0; i < m; i ++ ){
        cin >> k;
        num[k] ++;
    }
    int ans = num[n];
    for ( int i = 0; i <= 30000; i ++ ){
        for ( int j = 0; j <= 601; j ++ ){
            dp[i][j] = -9999999;
        }
    }
    dp[n][301] = num[n];
    for ( int i = n + 1; i <= 30000; i ++ ){
        for ( int j = 1; j <= 600; j ++ ){
            if ( i - ( j - 301 + n ) < 0 || i - ( j - 301 + n ) >= i )continue;
            dp[i][j] = max ( dp[i][j], dp[i-(j-301+n)][j-1] );
            dp[i][j] = max ( dp[i][j], dp[i-(j-301+n)][j]   );
            dp[i][j] = max ( dp[i][j], dp[i-(j-301+n)][j+1] );
            dp[i][j] += num[i];
            ans = max ( ans, dp[i][j] );
        }
    }
    cout << ans << endl;
}

D题：

题意：给出n个节点与m条有向边，判断最少需要边保证连通性不变

思路：最先想到的判断强联通，一个连通块中有若存在强联通分量，则需要的边数=节点数，若不存在强联通，则需要的边数=节点数-1；然后分别对每个连通块进行查询

PS：本来判强联通想通过tarjan染色，搞了好久搞不出来，看别人代码的时候看到了一种很神奇的代码，就拿来学习了下，果然大有收获。

下面是D题代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <vector>
#include <deque>
#include <list>
#include <cctype>
#include <algorithm>
#include <climits>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <ctime>
#define ll long long
#define ull unsigned long long
#define all(x) (x).begin(), (x).end()
#define clr(a, v) memset( a , v , sizeof(a) )
#define pb push_back
#define mp make_pair
#define read(f) freopen(f, "r", stdin)
#define write(f) freopen(f, "w", stdout)
using namespace std;
const double pi = acos(-1);
#define MAX_V 100005

int V, E;
int dfn[MAX_V], low[MAX_V];
bool instack[MAX_V];
int un[MAX_V];
int din;
int cnt;
int colorCnt = 0;
stack<int>s;
vector<int>Edge[MAX_V];
vector<int>g[MAX_V];
vector<int>ans;
int vis[MAX_V];

void init( int m )
{
    for ( int i = 0; i < MAX_V; i ++ ){
            Edge[i].clear();
    }
    for ( int i = 0; i <= MAX_V; i ++ ){
        un[i] = i;
    }
    ans.clear();
    while ( ! s.empty() )s.pop();
    clr ( dfn, 0 );
    clr ( low, 0 );
    clr ( vis, 0 );
    clr ( instack, 0 );
}

int find ( int x )
{
    if ( x == un[x] ){
        return x;
    }
    else{
        return un[x] = find( un[x] );
    }
}

void merge ( int x, int y )
{
    x = find( x );
    y = find( y );
    if ( x == y )return;
    un[x] = y;
    return;
}

int bfs ( int i )
{
    if ( Edge[i].size() >= 2 ){
        int cnt = 0;
        queue<int> q;
        for ( int j = 0; j < Edge[i].size(); j ++ ){
            if ( vis[Edge[i][j]] == 0 ) q.push( Edge[i][j] );
        }
        while ( ! q.empty() ){
            int x = q.front();
            q.pop();
            cnt ++;
            for ( int j = 0; j < g[x].size(); j ++ ){
                vis[g[x][j]] --;
                if ( vis[g[x][j]] == 0 ) q.push( g[x][j] );
            }
        }
        if ( cnt == Edge[i].size() ){
            return Edge[i].size() - 1;
        }
        else{
            return Edge[i].size();
        }
    }
    return 0;
}

int main()
{
    ios::sync_with_stdio( false );
    int m, n;
    while ( cin >> m >> n ){
        init ( m );
        int a, b;
        for ( int i = 0; i < n; i ++ ){
            cin >> a >> b;
            a --;
            b --;
            merge ( a, b );
            g[a].push_back( b );
            vis[b] ++;
        }
        for ( int i = 0; i < m; i ++ ){
            Edge[find(i)].push_back( i );
        }
        int ans = 0;
        for ( int i = 0; i < m; i ++ ){
            ans += bfs ( i );
        }
        cout << ans << endl;
    }
}

E题没做。

（未完待续）