大学生程序设计邀请赛（华东师范大学）

大学生程序设计邀请赛（华东师范大学）

比赛的时候写了3题，G还没A出来
AEFG/ 6.1补上B和C

A： 拼音魔法
就是模拟，输出拼音标音调后的。
注意a上标音调的话那个字符是占两位char的。
写的真的丑，但是语文太烂了好多不确定。。

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
char s[100][100][5] =
{
    {"a","e","i","o","u","ü"},
    {"ā","ē","ī","ō","ū","ǖ"},
    {"á","é","í","ó","ú","ǘ"},
    {"ǎ","ě","ǐ","ǒ","ǔ","ǚ"},
    {"à","è","ì","ò","ù","ǜ"}
};
int duiy[26] = {0};

char d[100];
int Find(char x,int dlen)
{
    for(int i = 0; i < dlen; i++)
    {
        if(d[i] == x)
        {
            if(x=='a' || x=='o' || x=='e' || x == 'v') return i;
            else if(x =='i')
            {
                if(d[i+1] == 'u')
                    return i+1;
                else return i;
            }
            else if(x == 'u')
            {
                if(d[i+1] == 'i')
                    return i+1;
                else return i;
            }
        }
    }
    return -1;
}
int print(int xid,int id)
{
    int dlen = strlen(d);
    for(int i = 0; i < dlen-1; i++)
    {
        if(i==xid)
            printf("%s",s[id][duiy[d[i]-'a']]);

        else if(d[i] == 'v')
            printf("ü");
        else
            printf("%c",d[i]);
    }
    printf("\n");
}
int main()
{
   int t;
   scanf("%d",&t);

   char ans[100];
   duiy['e'-'a'] = 1;
   duiy['i'-'a'] = 2;
   duiy['o'-'a'] = 3;
   duiy['u'-'a'] = 4;
   duiy['v'-'a'] = 5;
   for(int kase = 1; kase <= t; kase ++)
   {
       scanf("%s",d);
       int dlen = strlen(d);
       int flag = 0;
       int id;
       if(d[dlen -1] <='4' && d[dlen-1] >= '1')
        id = d[dlen-1]-'0';
       else id = 0;
       int aid = -1;
       for(int i = 0; i < dlen ;i++)
       {
           if(d[i] == 'v' && d[i+1] =='e')
                d[i] = 'u';
       }
       printf("Case %d: ",kase);
       if(id == 0)
       {
            for(int i = 0; i < dlen; i++)
           {
                if(d[i] == 'v')
                    printf("ü");
                else
                    printf("%c",d[i]);
           }
           printf("\n");
           continue;
       }
       else
       {
           int aid = Find('a',dlen);
           if(aid != -1)
                print(aid,id);
            else
            {
                int oid = Find('o',dlen);
                if(oid != -1)
                    print(oid,id);
                else
                {
                    int eid = Find('e',dlen);
                    if(eid != -1)
                        print(eid,id);
                    else
                    {
                        int iid = Find('i',dlen);
                        if(iid != -1)
                            print(iid,id);
                        else
                        {
                            int uid = Find('u',dlen);
                            if(uid != -1)
                            {
                                print(uid,id);
                            }
                            else
                            {
                                int vid = Find('v',dlen);
                                if(vid != -1)
                                    print(vid,id);

                            }
                        }
                    }
                }
            }
       }
    }
    return 0;
}

E 黑心啤酒厂
思路：如果n%i==0,直接一杯，不然结果就是i/gcd(n,i)
代码:

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
int gcd(int a,int b)
{
    if(b==0)
        return a;
    else return gcd(b,a%b);
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i = 2; i<=m ;i++)
    {
        if(n%i==0)
        {
            printf("1\n");
        }
        else
        {
            int k = gcd(n,i);
            int x = i/k;
            printf("%d\n",x);
        }
    }
    return 0;
}

F 丽娃河的狼人传说
思路：就是贪心，用r排序
代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1005;
struct node
{
    int l,r,m;
}d[maxn];
bool cmp(node a,node b)
{
    if(a.r == b.r)
        return a.l < b.l;
    else return a.r < b.r;
}
int vis[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    for(int kase = 1; kase <= t; kase ++)
    {
        memset(vis,0,sizeof(vis));
        int n,m,k;
        scanf("%d%d%d",&n,&m,&k);
        for(int i = 0; i < k ; i++)
        {
            int id;
            scanf("%d",&id);
            vis[id] = 1;
        }
        int flag = 0;
        for(int i = 0; i < m ; i++)
        {
            scanf("%d%d%d",&d[i].l,&d[i].r,&d[i].m);
            if(d[i].m > (d[i].r-d[i].l +1)) flag = 1;
        }
        printf("Case %d: ",kase);
        if(flag)
        {
            printf("-1\n"); continue;
        }
        sort(d,d+m,cmp);int ans = 0;
        for(int i = 0; i < m; i++)
        {
            int num = 0;

            for(int l = d[i].l; l <= d[i].r; l++)
            {
                if(vis[l] == 1)
                    num++;
            }
            if(num >= d[i].m)
                continue;
            else
            {
                int need = d[i].m-num;
                int x = 0;
                for(int r = d[i].r; r >= d[i].l; r--)
                {
                    if(vis[r] == 1) continue;
                    vis[r] = 1;
                    x++;
                    if(x == need) break;
                }
                ans += x;
            }
        }
        printf("%d\n",ans);

    }
    return 0;
}

G 铁路修复计划
思路： 二分+kruskal，二分k的值，求最大可以到多少，用kruskal来判断是否满足条件。
跑了14s…总共也就给了15s，哎
代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define met(a,b) memset(a,b,sizeof(a))
const int maxn = 100005;
const double eps = 1e-7;
int n,m;
LL k;
struct node
{
    int from,to;
    double w;int v;
    node(int x,int a,double b,int c)
    {
        from = x;to = a,w = b,v = c;
    }
    node(){}
}edge[maxn<<1];


node change[maxn<<1];
int f[maxn];

int edgenum = 0;
bool cmp(node a,node b)
{
   return a.w<b.w;
}
int getfather(int i)
{
    if(f[i] == i) return i;
    else return f[i] = getfather(f[i]);
}
int kruscal()
{
     sort(change,change+edgenum,cmp);

     int num = 0;
     double ans = 0;
     for(int i = 0; i <= n; i++)
        f[i] = i;
     for(int i = 0; i < m; i++)
     {
         int a = getfather(change[i].from),b = getfather(change[i].to);
         if(a!=b)
         {
             f[a] = b;
             double temp = change[i].w;
             ans += temp;
             num++;
             if(num == n-1)
                break;
         }
     }
     if(ans <= k) return 1;
     else return 0;

}

int solve()
{
    double ans = 0;
    double l = 1,r = 1e8;
    while(abs(l-r) > eps)
    {
        double mid = (l+r)/2.0;
        memcpy(change,edge,edgenum*sizeof(node));
        for(int i = 0;i < edgenum; i++)
        {
            if(change[i].v == 1)
                change[i].w = edge[i].w*mid;
        }
        int flag = kruscal();
        if(flag)    l = mid;
        else    r = mid;
    }
    printf("%.6f\n",l);
}

int main()
{
    for(int i = 0; i < maxn; i++)
    {
        f[i] = i;
    }
    scanf("%d%d%lld",&n,&m,&k);
    for(int i = 0; i < m; i++)
    {
        int a,b,d;
        double c;
        scanf("%d%d%lf%d",&a,&b,&c,&d);
        edge[edgenum++] = node(a,b,c,d);
    }
    solve();
    return 0;
}

B 分词
思路：
其实就是考dp，当时没怎么看这道题。。
用dp[i]来表示以i为结尾的最大值，dp[i] = dp[k]+val[thisword];
不过这题改了挺长时间的，注意细节啊。 边界值老是出错。还有刚开始题意都没读懂就开始写了是个什么鬼
代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <cmath>
using namespace std;
map<string,double> value;
const int inf = 0x3f3f3f3f;
char data1[5010],data[5010],s[5010];
double dp[5010];
char test[50];
int pre[5010],vis[5010];

const double eps = 1e-9;
int tolow(int slen)
{
    for(int i = 0; i < slen; i++)
        if(isupper(s[i])) s[i] = s[i]+32;
    return 0;
}
void tolow2(int dlen)
{
    for(int i = 0; i < dlen; i++){
        if(isupper(data1[i])) data[i] = data1[i]+32;
        else data[i] = data1[i];
    }
    data[dlen] = '\0';
}
int main()
{
    int n,m;
    scanf("%d",&n);

    double v;
    for(int i = 0; i < n ; i++)
    {
        scanf("%s%lf",s,&v);
        if(strlen(s) > 30) value[s] = 0;
        else
        {
            int slen = strlen(s);
            tolow(slen);
            value[s] = slen*slen*log(v);
        }
    }
    scanf("%d",&m);
    for(int i = 0; i < m ; i++)
    {
        scanf("%s",data1);
        int dlen = strlen(data1),ansi = -1;

        tolow2(dlen);
        double ans = 0;

        for(int j = 0; j < dlen; j++) dp[j] = 0, pre[j] = -1,vis[j] = 0;
        for(int j = 0; j < dlen; j++)
        {
            double res = 0,th;
            int k,num;
            k = j-30 > -1 ? j-30 : -1;
            for(; k < j; k++)
            {
                num = 0,th = 0;
                for(int x = k+1; x <= j; x++)
                    test[num++] = data[x];
                test[num] = '\0';
                if(value.count(test)) th = value[test];
                if(res < dp[k] + th)
                {
                    pre[j] = k;
                    res = dp[k]+th;
                }
            }
            dp[j] = res;
            if(ans < res)   ans = res,ansi = j;            
        }
        printf("%.6f\n",ans);
       // for(int j = 0; j < dlen; j++)cout<<dp[j]<<endl;
        int k = ansi;
        while(k != -1)
        {
            vis[k] = 1;
            k = pre[k];
        }
        for(int j = 0; j < dlen; j++)
            vis[j]&&j != dlen-1?printf("%c ",data1[j]) : printf("%c",data1[j]);
        printf("\n");
    }
    return 0;
}

C 袋鼠妈妈
思路： 看了题解才写出来，而且由于写的时候dfs姿势不好还tle，这个故事告诉我们有些大佬的暴力其实跟我们的暴力根本就不一样（没什么联系）。
对这题，只要卡，到这个位置 只能有一条路过来，基本就是正确的结果，把其余位置全部造墙，只有这条路能走，感觉就是瞎搜。。。但是莫名其妙啊，我把way[][]的顺序换了一波才成功ac，不知道哪里出了问题。。

代码：

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

using namespace std;
int n,m,x,y,k;

int vis[10][10];
int way[4][2] = {0,1,0,-1,-1,0,1,0};
int dfs(int nx,int ny,int s)
{
    if(s >= k && nx == x && ny == y)
    {
        for(int i = 0; i < n; i ++)
        {
            for(int j = 0; j < m ; j++)
                vis[i][j] == 1?printf(".") : printf("*");
            printf("\n");
        }
        return true;
    }

    for(int i = 0; i < 4 ; i++)
    {
        int cx = nx + way[i][0], cy = ny + way[i][1];
        if(cx < 0 || cx >= n || cy < 0 || cy >= m || vis[cx][cy]) continue;
        int sum = 0;
        for(int j = 0; j < 4; j++)
            if(vis[cx+way[j][0] ][cy + way[j][1]]) sum++;

        if(sum >= 2) continue;
        vis[cx][cy] = 1;
        if(dfs(cx,cy,s+1)) return true;//找到了就不需要找了，不然多找也要费时间。
        vis[cx][cy] = 0;
    }
    return false;
}
int main()
{
    scanf("%d%d%d%d%d",&n,&m,&x,&y,&k);
    vis[0][0] = 1;
    x--,y--;
    dfs(0,0,0);
    return 0;
}