UESTC 2014 Summer Training #10 Div.2

B.Race to 1 UVA 11762

　　第一次接触概率dp,完全没想到是dp...没想到能递推出来0 0

　　首先需要知道 总的期望=每件事的期望×每件事发生的概率

　　然后可以根据这个来写递推公式,也是dp?

　　假设不小于x的质数有m个,x的质因子有n个(种 更确切),那么在求X的期望时,可以考虑由他能转移过去的状态转移,X,X/pi p是x的质因子

　　所以不难得出E(x) = 1+(m-n)/mE(X)+1/msigmaE(X/pi)　　注意会加1,因为X转移到后面的情况,就多了一步

//Updated:　　每个case跑完了,f数组不用memset...会快10倍!uva数据的情况下

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>

using namespace std;

const int maxn = 1000000+50;
const double eps = 1e-10;

int T, N, tot;
int prime[maxn];
double f[maxn];
bool flag[maxn], done[maxn];

void getPrime()
{
    for(int i = 2; i < maxn; i++) {
        if(!flag[i])    prime[tot++] = i;
        for(int j = 0; j < tot && prime[j]*i < maxn; j++) {
            flag[i*prime[j]] = true;
            if(i % prime[j] == 0)    break;
        }
    }
}

double dp(int x)
{
    if(done[x])    return f[x];
    done[x] = true;
    int n, m;
    n = m = 0;
    for(int i = 0; i < tot && prime[i] <= x; i++) {
        n++;
        if(x % prime[i] == 0) {
            m++;
            f[x] += dp(x/prime[i]);
        }
    }
//    if(x == N)    cout << n << ' ' << m << endl;
    return f[x] = (f[x]+n)/m;
}

int main()
{
#ifdef LOCAL
    freopen("B.in", "r", stdin);
#endif
    getPrime();
    scanf("%d", &T);
    for(int t = 1; t <= T; t++) {
        scanf("%d", &N);
        memset(done, 0, sizeof(done));
        memset(f, 0, sizeof(f));
        f[1] = 0.0;    done[1] = true;
        dp(N);
        printf("Case %d: %.10lf
", t, f[N]);
    }
    return 0;
}

D.Jumping Mario UVA 11764

#include <iostream>
#include <cstdio>
#include <cstdlib>

using namespace std;

int T, n, now, high, low;

int main()
{
    cin >> T;
    for(int t = 1; t <= T; t++) {
        high = 0;
        low = 0;
        cin >> n;
        for(int i = 0; i < n; i++) {
            int x;
            cin >> x;
            if(i == 0) {
                now = x;
                continue;
            }
            if(x > now)    high++;
            else if(x < now)    low++;
            now = x;
        }
        cout << "Case " << t << ": " << high << ' ' << low << endl;
    }
    return 0;
}

J.Lighting Away UVA 11770

　　第一反应的确是强连通分量...缩点...但是我不太熟...所以就去yydfs的方法,可是智商不够用...最后没调试出来

　　正解就是缩点染色判断新的图入读为0的点

　　调试了很久很久...各种小错误都发现了,最后倒在了出栈忘记加大括号了...

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <stack>
#include <cstring>

using namespace std;

const int maxn = 100000+50;

int T, n, m, idx, tot, cnt;
int head[maxn], next[maxn], edge[maxn], dfn[maxn], low[maxn], in[maxn], type[maxn];
bool instack[maxn], visit[maxn];
stack<int> st;

void add(int u, int v)
{
    edge[tot] = v;
    next[tot] = head[u];
    head[u] = tot++;
}

void tarjan(int u)
{
    dfn[u] = low[u] = ++idx;
    st.push(u);
    instack[u] = true;
    visit[u] = true;
    for(int e = head[u]; e != -1; e = next[e]) {
        int v = edge[e];
        if(!visit[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }        
        else if(instack[v])
            low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u]) {
        cnt++;
        while(!st.empty())
        {
            int now=st.top();    st.pop();
            instack[now]=0;
            type[now]=cnt;
            if(now == u) break;
        }
    }
}

int main()
{
#ifdef LOCAL
    freopen("J.in", "r", stdin);
#endif
    scanf("%d", &T);
    for(int t = 1; t <= T; t++) {
        scanf("%d%d", &n, &m);
        memset(next, -1, sizeof(next));
        memset(head, -1, sizeof(head));
        memset(in, 0, sizeof(in));
        memset(instack, 0, sizeof(instack));
        memset(visit, 0, sizeof(visit));
        tot = 0;    idx = 0;    cnt = 0;
        for(int i = 0; i < m; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            add(u, v);
        }
//        for(int i = 1; i <= n; i++)
//            if(!visit[i] && in[i] == 0)    tarjan(i);

        for(int i = 1; i <= n; i++) 
            if(!visit[i])    tarjan(i);
        int ans = 0;
        for(int i = 1; i <= n; i++)
            for(int e = head[i]; e != -1; e = next[e]) {
                if (type[i] != type[edge[e]]) in[type[edge[e]]]++;
            }
        for (int i = 1;i <= cnt;i++)
            if (in[i] == 0) ans++;
//        for(int i = 1; i <= n; i++)
//            if(in[i] == 0)    cnt++;
        printf("Case %d: %d
", t, ans);

    }
    return 0;
}

　　这场比赛做的时候只出了道水题...然后J和B是真的实力不够...J就是后来搞也有很多迷糊的地方,现在算是搞懂了