Regionals 2012 :: Asia

水 B Wedding of Sultan

题意：求每个点的度数

分析：可以在，每个字母的的两个端点里求出的的出度，那么除了起点外其他点还有一个入度，再+1

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 13:22:03
* File Name     :B.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int vis[33];
char s[N];
int p1[33], p2[33];
int ans[33];

int main(void)    {
    int T, cas = 0; scanf ("%d", &T);
    while (T--) {
        scanf ("%s", &s);
        memset (ans, -1, sizeof (ans));
        memset (vis, 0, sizeof (vis));
        memset (p1, -1, sizeof (p1));
        memset (p2, -1, sizeof (p2));
        int len = strlen (s);
        for (int i=0; i<len; ++i)    {
            if (ans[s[i]-'A'] == -1)    ans[s[i]-'A'] = 0;
            if (p1[s[i]-'A'] == -1) p1[s[i]-'A'] = i;
            else    p2[s[i]-'A'] = i;
        }
        for (int i=0; i<len; ++i)   {
            if (vis[s[i]-'A'] == 1)  continue;
            int j = p1[s[i]-'A'] + 1;
            if (j == p2[s[i]-'A'])  {
                //ans[s[i]-'A'] = 1;  vis[s[i]-'A'] = 1;
                continue;
            }
            while (j < p2[s[i]-'A'])   {
                ans[s[i]-'A']++;
                j = p2[s[j]-'A'] + 1;
            }
            vis[s[i]-'A'] = 1;
        }
        for (int i=0; i<26; ++i)    ans[i]++;
        ans[s[0]-'A']--;

        printf ("Case %d
", ++cas);
        for (int i=0; i<26; ++i)    {
            if (ans[i] <= 0)    continue;
            printf ("%c = %d
", 'A' + i, ans[i]);
        }
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}

水 C Memory Overflow

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 12:55:22
* File Name     :C.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 5e2 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
char s[N];

int main(void)    {
    int T, cas = 0;  scanf ("%d", &T);
    while (T--) {
        int n, k;   scanf ("%d%d", &n, &k);
        scanf ("%s", &s);
        int ans = 0;
        for (int i=0; i<n; ++i) {
            bool flag = false;
            for (int j=max (0, i-k); j<i; ++j)  {
                if (s[j] == s[i])   {
                    flag = true;    break;
                }
            }
            if (flag)   ans++;
        }
        printf ("Case %d: %d
", ++cas, ans);
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}

暴力/高斯消元 E Poker End Games

只会暴力

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 18:34:27
* File Name     :E.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);

int main(void)    {
    int T, cas = 0; scanf ("%d", &T);
    while (T--) {
        int a, b;   scanf ("%d%d", &a, &b);
        double p = 1.0, round = 0, win = 0;
        int n = 1e6;
        for (int i=1; i<=n; ++i)    {
            if (a == b) {
                round += p * i;
                win += p * 0.5;
                break;
            }
            if (a > b)  {
                a -= b; b += b;
                round += p * i * 0.5;
                win += p * 0.5;
            }
            else if (a < b) {
                b -= a; a += a;
                round += p * i * 0.5;
            }
            p *= 0.5;
        }
        printf ("Case %d: %.6f %.6f
", ++cas, round, win);
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}

　　

水 F Overlapping Characters

题意：判断能否根据某个点是'*'来区别出形状，必须是‘*'，其他的都是'.'才行。

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 14:03:02
* File Name     :F.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
char s[40][20][50];
char ss[40];
int id[40];
int ans[20];

int f(char ch) {
    if (ch >= 'A' && ch <= 'Z') return ch - 'A';
    else    return 26 + (ch - '0');
}

int main(void)    {
    int n, m, cas = 0;   scanf ("%d%d", &n, &m);
    scanf ("%s", &ss);
    //printf ("%s
", ss);
    memset (id, -1, sizeof (id));
    for (int i=0; i<n; ++i) {
        id[f (ss[i])] = i;
        for (int j=0; j<17; ++j)    {
            scanf ("%s", s[i][j]);
        }
        /*
        for (int j=0; j<17; ++j)    {
            printf ("%s
", s[i][j]);
        }
        */
    }

    for (int i=0; i<m; ++i) {
        scanf ("%s", &ss);
        int len = strlen (ss);
        for (int j=0; j<len; ++j)   {
            ans[j] = 0;
            if (id[f (ss[j])] == -1)    continue;
            for (int k=0; k<16 && !ans[j]; ++k)    {       //row
                for (int l=0; l<43 && !ans[j]; ++l)    {       //col
                    bool flag = true;
                    if (s[id[f (ss[j])]][k][l] != '*')    continue;
                    for (int ii=0; ii<len; ++ii)    {
                        if (ii == j)    continue;
                        if (id[f (ss[ii])] == -1)   continue;
                        if (s[id[f (ss[ii])]][k][l] != '.')  {
                            flag = false;   break;
                        }
                    }
                    if (flag)   ans[j] = 1;
                }
            }
        }
        printf ("Query %d: ", ++cas);
        for (int i=0; i<len; ++i)   {
            printf ("%c", ans[i] ? 'Y' : 'N');
        }
        puts ("");
    }


   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}

　　

DP I Learning Vector

题意：给了一些向量，问选出一些与x坐标轴组成的多变形面积最大

分析：开始以为贪心，因为很难想DP把之前的选的向量保存起来。看题解才知道，dp[i][j][h] 表示前i个选择了j个，最后高度为h组成的面积，那么状态转移时不需要知道之前的长度，用梯形面积公式累加多出来的面积就行了。另外，向量应该先极角排序。

/************************************************
* Author        :Running_Time
* Created Time  :2015/11/4 星期三 15:47:19
* File Name     :I_2.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x)  {       //三态函数，减少精度问题
    if (fabs (x) < EPS) return 0;
    else    return x < 0 ? -1 : 1;
}
struct Vec    {
    int x, y;
    Vec () {}
    Vec (int x, int y) : x (x), y (y) {}
}vs[55];
bool cmp(Vec A, Vec B)    {
    return dcmp (A.x * B.y - A.y * B.x) < 0;
}

int dp[55][55][2550];

int main(void)    {
    int T, cas = 0; scanf ("%d", &T);
    while (T--) {
        int n, k;   scanf ("%d%d", &n, &k);
        int H = 0;
        for (int x, y, i=1; i<=n; ++i) {
            scanf ("%d%d", &vs[i].x, &vs[i].y);
            H += vs[i].y;
        }
        sort (vs+1, vs+1+n, cmp);
        memset (dp, -1, sizeof (dp));
        dp[0][0][0] = 0;
        for (int i=0; i<n; ++i)    {
            for (int j=0; j<=i && j<=k; ++j)    {
                for (int h=0; h<=H; ++h)    {
                    if (dp[i][j][h] == -1)  continue;
                    dp[i+1][j][h] = max (dp[i+1][j][h], dp[i][j][h]);
                    if (j < k)  {
                        int hh = h + vs[i+1].y;
                        dp[i+1][j+1][hh] = max (dp[i+1][j+1][hh], dp[i][j][h] + (h + hh) * vs[i+1].x);
                    }
                }
            }
        }
        
        int ans = 0;
        for (int i=0; i<=H; ++i)    if (ans < dp[n][k][i])  ans = dp[n][k][i];
        printf ("Case %d: %d
", ++cas, ans);
    }

   //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.
";

    return 0;
}