1140

题目大意： 题意说的很复杂。其实就是按序统计前一个字符串有连续几个相同的字符.

#include<iostream>
#include<string>
using namespace std;

int n;
string d;

void change(string s, int count){
    if(count == n){
        cout<<s<<endl;
        return;
    }
    string temp = "";
    int i, j, num;
    for(i = 0; i < s.length(); i++){
        num = 0;
        temp += s[i];
        for(j=i; j<s.length() ;j++){
            if(s[j]!=s[i]){
                break;
            }else{
                num++;
            }
        }
        i = j-1;
        temp += num + '0';
    }
    change(temp, count+1);
}
int main(){
    cin>>d>>n;
    change(d,1);
    return 0;
} 

1141 STL

题目大意：

• 输入id ,分数， 学校。
• 第一行输出总共有多少个学校
• 排序，计算加权分数，将分数从高到低排序,假如分数相同，则按照队伍数量从小到大排列 ,再按学校名称的字母表排序 。
• 打印 名词，学校，加权总分 ，以及每个学校有的队伍

#include<iostream>
#include<string>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;

typedef struct node{
    string school;
    int num_tees;//队伍数量
    int score;//加权分数 
}node; 

bool cmp(node a, node b){
    if(a.score == b.score){
        if(a.num_tees == b.num_tees){
            return a.school < b.school;
        }
        return a.num_tees < b.num_tees;
    } 
    return a.score > b.score;
}

int main(){
    int n; // n<=100000
    cin>>n;
    map<string, double> all_score;
    map<string, int> all_tees;
    
    for(int i=0;i<n;i++){
        string id, school;
        double score;
        cin>>id>>score>>school;
        //转换成小写 
        transform(school.begin(), school.end(), school.begin(), ::tolower);
        if(id[0] == 'B'){
            score = score / 1.5;
        }else if(id[0] == 'T'){
            score = score * 1.5;
        }
        all_tees[school]++;
        all_score[school] += score;
    }
    
    map<string, double>::iterator it1;
    map<string, int>::iterator it2;
    vector<node> v;
    for(it1 = all_score.begin(), it2 = all_tees.begin(); it1!=all_score.end(); it1++, it2++){
        node temp;
        temp.school = it1->first;
        temp.score = int(it1->second);
        temp.num_tees = it2->second; 
        v.push_back(temp);
    }
    sort(v.begin(), v.end(), cmp); 
    printf("%d\n",v.size());
    int rank =1, init_score = v[0].score;
    cout<<rank<<" "<<v[0].school<<" "<<v[0].score<<" "<<v[0].num_tees<<endl;
    for(int i=1;i<v.size();i++){
        if(v[i].score != init_score){
            rank = i+1;
            init_score = v[i].score;
        }
        cout<<rank<<" "<<v[i].school<<" "<<v[i].score<<" "<<v[i].num_tees<<endl;
    }
    return 0;
}

1142 图论

题目大意： 让你判断是clique, maximal clique 还是什么都不是。

• clique：一个图节点的子集，节点之间都是相邻的
• maximal clique：不能增加节点的clique

#include<stdio.h>
#include<vector>
using namespace std;

int main(){
    int n, m;
    scanf("%d %d", &n, &m);
    int e[205][205] = {0}; 
    for(int i=0;i<m;i++){
        int a,b;
        scanf("%d %d", &a, &b);
        e[a][b] = e[b][a] = 1;
    }
    int k;
    scanf("%d", &k);
    for(int i=0;i<k;i++){
        int a;
        scanf("%d", &a);
        vector<int> v(a);
        vector<int> table(n+1);
        for(int j=0;j<a;j++){
            scanf("%d", &v[j]);
            table[v[j]] = 1;
        }
        //先看是否都相邻
        bool flag1 = true;
        for(int p=0;p<a && flag1;p++){
            for(int q=p+1;q<a;q++){
                if(e[v[p]][v[q]] == 0){
                    flag1 = false;
                    break;
                }
            }
        }
        if(flag1 == false){
            printf("Not a Clique\n");
            continue;
        }
        //再看是否能加进去点 
        bool flag2 = true;
        for(int p=1;p<=n;p++){
            if(table[p]==0){
                int q=0;
                for(;q<a;q++){
                    if(e[p][v[q]]==0){
                        break;
                    }
                }
                if(q==a){
                    flag2 = false;
                    break;
                }
            }
        }
        if(flag2 == false){
            printf("Not Maximal\n");
            continue;
        }
        printf("Yes\n");
    }
    return 0;
}

1143 树

题目大意： 在二叉搜索树中找两个结点的最近祖先 -> 转化成知道先序、中序（从小到大顺序）求最近祖先.

#include<stdio.h>
#include<vector>
#include<map>
using namespace std;

int main(){
    int m,n;
    scanf("%d %d", &m, &n);
    vector<int> preorder(n);
    map<int, bool> mp;
    for(int i=0;i<n;i++){
        scanf("%d", &preorder[i]);
        mp[preorder[i]] = true;
    }
    
    for(int i=0;i<m;i++){
        int u,v;
        scanf("%d %d", &u, &v);
        if(mp[u] == false && mp[v] == true){
            printf("ERROR: %d is not found.\n", u);
            continue;
        }else if(mp[v] == false && mp[u] == true){
            printf("ERROR: %d is not found.\n", v);
            continue;
        }else if(mp[v] == false && mp[u] == false){
            printf("ERROR: %d and %d are not found.\n", u, v);
            continue;
        }
        int anc;
        for(int j=0;j<n;j++){
            anc = preorder[j];
            if( (anc >=u && anc <=v ) || (anc >=v && anc <=u )){
                break;
            }
        }
        if(anc == u){
            printf("%d is an ancestor of %d.\n", u, v);
        }else if(anc == v){
            printf("%d is an ancestor of %d.\n", v, u);
        }else{
            printf("LCA of %d and %d is %d.\n", u, v, anc);
        }
    }
    return 0;
}