BestCoder Round #25 A，B

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挺好的一场比赛，完全被自己的智商给碾压了啊。。。都是泪啊。。

A，判断有向图中是否有环，数据很小简单粗暴的暴力算法可解啊。暴力枚举有关系两个点判断反向是否可以找到，如果可以就说明有环。。。

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

using namespace std;

const int maxn = 210;

int vis[maxn];
int mp[maxn][maxn];

vector<int> g[maxn];

int flag;


void dfs(int x, int y)
{
    if(x == y)
    {
        flag = 1;
        return;
    }
    if(vis[x]) return;
    vis[x] = 1;
    int n = g[x].size();
    for(int i = 0; i < n; i++) dfs(g[x][i], y);
}

int main()
{
    int n;
    int m;
    while(cin >>n>>m)
    {
        int x, y;
        flag = 0;
        memset(mp, 0, sizeof(mp));
        for(int i = 0; i <= n; i++) g[i].clear();
        for(int i = 0; i < m; i++)
        {
            scanf("%d %d",&x, &y);
            mp[x][y] = 1;
            g[x].push_back(y);
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                if(!mp[i][j]) continue;
                memset(vis, 0, sizeof(vis));
                dfs(j, i);
                if(flag) break;
            }
            if(flag) break;
        }
        if(flag) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
    }
}

题解里面解释的很清楚了啊。。不再瞎说了啊。

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007

using namespace std;

const int maxn = 210;

LL dp[maxn][maxn];
LL cnk[maxn][maxn];

void init()
{
    memset(cnk, 0, sizeof(cnk));
    cnk[0][0] = 1LL;
    for(int i = 1; i < 51; i++) cnk[i][0] = cnk[i][i] = 1LL;
    for(int i = 2; i < 51; i++)
        for(int j = 1; j < i; j++) cnk[i][j] = (cnk[i-1][j-1] + cnk[i-1][j])%mod;
}



int main()
{
    int n, m;
    init();
    while(~scanf("%d %d",&n, &m))
    {
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1LL;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j <= m; j++)
            {
                for(int k = 1; k <= m; k++)
                {
                    for(int t = m; t >= 0; t--)
                    {
                        if(m-j < 0 || k-t < 0) continue;
                        dp[i+1][j+t] += (dp[i][j]*cnk[m-j][t]%mod)*cnk[j][k-t]%mod;
                        dp[i+1][j+t] %= mod;
                    }
                }
            }
        }
        cout<<dp[n][m]%mod<<endl;
    }
}

BestCoder Round #25 A，B

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原文地址：http://blog.csdn.net/xu12110501127/article/details/42386587