CUGBACM Codeforces Tranning 3 题解

链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=62515#overview

描述：第三场CF训练了，这次做的挺搞笑的，我记得这是内天持续训练九个小时中的最后两个小时，想想也是蛮拼的。

题解：

A.Triangle

题意：给四个边，如果能组成判断能不能从其中找三条边组成三角形，不就再判断能不能三条边首尾相接组成一个线段。

思路：三角形判断条件，两条边之和大于第三条边，两边之和等于第三条边就是线段。

代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <ctype.h>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define eps 1e-8
#define INF 0x7fffffff
#define maxn 10005
#define PI acos(-1.0)
#define seed 31//131,1313
#define LOCAL
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
int main()
{
    int a[5];
    for(int i=0;i<4;i++)
        scanf("%d",&a[i]);
    sort(a,a+4);
    if(a[0]+a[1]>a[2]||a[1]+a[2]>a[3])
        puts("TRIANGLE");
    else if(a[0]+a[1]==a[2]||a[1]+a[2]==a[3]||a[0]+a[2]==a[3])
        puts("SEGMENT");
    else puts("IMPOSSIBLE");
    return 0;
}

B.Alice, Bob and Chocolate

题意：两个人，一个从左边开始吃巧克力，一个右边开始吃巧克力，两个人吃的速度是一样的，如果两人同时开始吃同一块巧克力，右边人放弃。问两边人各吃多少个巧克力。

思路：记每个个每个巧克力吃的时间，比一下就行。

代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <ctype.h>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define eps 1e-8
#define INF 0x7fffffff
#define maxn 10005
#define PI acos(-1.0)
#define seed 31//131,1313
#define LOCAL
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
int main()
{
    int tot;
    scanf("%d",&tot);
    int aa[100005];
    int l[100005],r[100005];
    int sum = 0;
    l[0]=0;
    for(int i=0;i<tot;i++)
    {
        scanf("%d",&aa[i]);
    }
    for(int i=1;i<tot;i++)
        l[i]=l[i-1]+aa[i-1];
    r[tot-1]=0;
    for(int i=tot-2;i>=0;i--)
        r[i]=r[i+1]+aa[i+1];
    int Alice = 0,Bob = 0;
    for(int i=0;i<tot;i++)
    {
        if(l[i]<=r[i])
        Alice++;
    else Bob++;
    }
    cout<<Alice<<" "<<Bob<<endl;
}

C.President‘s
Office

题意：给一个n*m的图，里面用大写字母表示桌子，找到和给出的字母相邻的字母的种类数之和。

思路：对于每个存在的字母找一下上，下，左，右即可。

代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <ctype.h>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define eps 1e-8
#define INF 0x7fffffff
#define maxn 10005
#define PI acos(-1.0)
#define seed 31//131,1313
#define LOCAL
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
char ss[105][105];
bool vis[60];
int main()
{
    int row,col;
    char cap[5];
    scanf("%d%d%s",&row,&col,cap);
    for(int i=0;i<row;i++)
        scanf("%s",ss[i]);
    for(int i=0;i<row;i++)
    {
        for(int j=0;j<col;j++)
        {
            if(ss[i][j]==cap[0])
            {
                if(i>0)
                    if(ss[i-1][j]!=cap[0]&&ss[i-1][j]>='A')
                        vis[ss[i-1][j]-'A']=1;
                if(j>0)
                    if(ss[i][j-1]!=cap[0]&&ss[i][j-1]>='A')
                        vis[ss[i][j-1]-'A']=1;
                if(i<row)
                    if(ss[i+1][j]!=cap[0]&&ss[i+1][j]>='A')
                        vis[ss[i+1][j]-'A']=1;
                if(j<col)
                    if(ss[i][j+1]!=cap[0]&&ss[i][j+1]>='A')
                        vis[ss[i][j+1]-'A']=1;
            }
        }
    }
    int ans = 0;
    for(int i=0;i<26;i++)
        if(vis[i])
        ans++;
    printf("%d\n",ans);
    return 0;
}

D.Longest Regular Bracket Sequence

题意：给一个长的字符串，其中都是"("和")"，问最长的合理括号匹配子串是多长，并且找出该长度的子串有多少个。

思路：比赛的时候，了。是写DP搞了好久都搞不出来，加上脑子糊里糊涂的，就GG了。思路是这样的，对于每个")"，如果它左边的是合理的匹配，并且找到它左边的合理匹配的第一个"("的左端也是"("，那么它的匹配数是左边的匹配数+2，即dp[i]=dp[i-1]+2，并且如果找到的合理匹配的左端还存在合理匹配，那么它的合理匹配长度还要加上前面合理匹配的长度。

代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <ctype.h>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define eps 1e-8
#define INF 0x7fffffff
#define maxn 10005
#define PI acos(-1.0)
#define seed 31//131,1313
#define LOCAL
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
int dp[1000005];
char ss[1000005];
int main()
{
    int i , time = 1,ans = 0;
    scanf("%s",ss);
    memset(dp,0,sizeof(dp));
    for(i=1; ss[i]!='\0'; i++)
    {
        if(i-dp[i-1]-1>=0&&ss[i]==')'&&ss[i-dp[i-1]-1]=='(')
        {
            dp[i]=dp[i-1]+2;
            if(i-dp[i-1]-2>=0)
                dp[i]+=dp[i-dp[i-1]-2];
        }
        if(dp[i]>ans)
        {
            ans = dp[i];
            time = 1;
        }
        else if(dp[i]==ans&&ans!=0)
        {
            time++;
        }
    }
    printf("%d %d\n",ans,time);
}

E.Exposition

题意：给出N本书，n<=10^5，这n本书按出版时间给出的，给出了每本书的高度hi。并且给出一个k，k<=10^6，要求在n本书选择其中连续的若干本书，其中最高的高度比最高的高度不能大于k，问最多可以选择多少本书。

思路：二分+RMQ。从左至右依次选择每本书作为起点，二分来选择符合条件的终点来保证以该本书为起点的长度最长，对于每次的终点，找到起点终点之间的最高高度和最低高度，如果符合条件，那么终点右移，否则终点左移。查询最高高度和最低高度的方式是ST表，O(NlogN)的预处理，然后O(1)的查询，二分的过程中的复杂度也是O(NlogN)。

代码：

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <ctype.h>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#define eps 1e-8
#define INF 0x7fffffff
#define maxn 100005
#define PI acos(-1.0)
#define seed 31//131,1313
#define LOCAL
typedef long long LL;
typedef unsigned long long ULL;
using namespace std;
int stTable_min[maxn][32],stTable_max[maxn][32];
int preLog2[maxn];
int h[maxn];
int Left[maxn],Right[maxn];
int ans = 0,times = 0;
void st_prepare(int n,int *array)
{
    preLog2[1]=0;
    for(int i=2; i<=n; i++)
    {
        preLog2[i]=preLog2[i-1];
        if((1<<preLog2[i]+1)==i)
            preLog2[i]++;
    }
    for(int i=n-1; i>=0; i--)
    {
        stTable_min[i][0]=array[i];
        stTable_max[i][0]=array[i];
        for(int j=1; (i+(1<<j)-1)<n; j++)
        {
            stTable_min[i][j]=min(stTable_min[i][j-1],stTable_min[i+(1<<j-1)][j-1]);
            stTable_max[i][j]=max(stTable_max[i][j-1],stTable_max[i+(1<<j-1)][j-1]);
        }
    }
    return ;
}
int query_sub(int l,int r)
{
    int len=r-l+1,k=preLog2[len];
    return max(stTable_max[l][k],stTable_max[r-(1<<k)+1][k])-min(stTable_min[l][k],stTable_min[r-(1<<k)+1][k]);
}
int main()
{
    int all = 0;
    int tot,k;
    scanf("%d%d",&tot,&k);
    for(int i=0;i<tot;i++)
        scanf("%d",&h[i]);
    st_prepare(tot,h);
    for(int i=0;i<tot;i++)
    {
        int l = i , r = tot;
        while(r>l+1)
        {
            int mid = (l + r) / 2;
            if(query_sub(i,mid)>k)
                r=mid;
            else l=mid;
        }
        if(l-i+1>ans)
        {
            ans=l-i+1;
            times=1;
            Left[0]=i;
            Right[0]=l;
        }
        else if(l-i+1==ans)
        {
            Left[times]=i;
            Right[times++]=l;
        }
    }
    printf("%d %d\n",ans,times);
    for(int i=0;i<times;i++)
        printf("%d %d\n",Left[i]+1,Right[i]+1);
    return 0;
}
时间： 2024-10-10 17:34:58

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