PTA 1140 1141 1142 1143

1140 Look-and-say Sequence

思路：模拟

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
int d,n;
const int maxn = 10010;
int cnt[11];
vector<int> a;
vector<int> b;

int main(){
	scanf("%d%d",&d,&n);
	a.push_back(d); 
	if(n == 1) {
		printf("%lld",d);
		return 0;
	}
	for(int i=2;i<=n;i++){
		int j = 0;
		b.clear();
		int len = a.size();
		while(j < len){
			int curv = a[j];
			int cnt = 0;
			while(j<len && a[j] == curv){
				cnt++;
				j++;
			}
			b.push_back(curv);
			b.push_back(cnt);
		}
		a.clear();
		for(int j=0;j<b.size();j++) a.push_back(b[j]);
	}
	for(int i=0;i<a.size();i++){
		printf("%d",a[i]);
	}
	return 0;
}

1141 PAT Ranking of Institutions

思路：结构体排序，map统计查询

#include<bits/stdc++.h>
#include<algorithm>
#include<string>
using namespace std;

int n;
const int maxn = 1e5+10;
struct node{
	string id;
	double score;
	string cap;
};

struct node stu[maxn];
map<string,int> mp;
map<string,double> scores;

struct capNode{
	string cap;
	int score;
	int nums;
	capNode(string ss,double cc,int num){
		cap = ss;
		score = cc;
		nums = num;
	}
};

vector<capNode> vec;

bool cmp(capNode x,capNode y){
	if(x.score == y.score) {
		if(x.nums == y.nums) return x.cap < y.cap;
		return x.nums < y.nums;
	}
	return x.score > y.score;
}

int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		string curCap;
		cin>>stu[i].id>>stu[i].score>>curCap;
		transform(curCap.begin(),curCap.end(),curCap.begin(),::tolower);
		stu[i].cap = curCap;
		mp[stu[i].cap]++;
		if(stu[i].id[0] == 'T'){
			scores[stu[i].cap] += stu[i].score*1.5;
		}else if(stu[i].id[0] == 'A'){
			scores[stu[i].cap] += stu[i].score;
		}else{
			scores[stu[i].cap] += stu[i].score*1.0/1.5;
		}
	}
	map<string,int>::iterator it = mp.begin();
	while(it!=mp.end()){
		string name = it->first;
		int num = it->second;
		int sc = (int)scores[name];
		vec.push_back(capNode(name,sc,num));
		it++;
	}
	sort(vec.begin(),vec.end(),cmp);
	int len = vec.size();
	printf("%d
",len);
	if(len > 0){
		int lastScore = vec[0].score;
		int rank = 1;
		for(int i=0;i<len;i++){
			if(vec[i].score == lastScore){
				cout<<rank<<" "<<vec[i].cap<<" "<<vec[i].score<<" "<<vec[i].nums<<endl;
			}else{
				rank = i+1;
				lastScore = vec[i].score;
				cout<<rank<<" "<<vec[i].cap<<" "<<vec[i].score<<" "<<vec[i].nums<<endl;
			}
		}
	}
	return 0;
} 

1142 Maximal Clique

思路：图论，判断是否同一“集合”，按题目要求判断是否有边

#include<bits/stdc++.h>
using namespace std;

const int maxn = 210;
int n,ne;
int m;
int g[maxn][maxn];
int a[maxn];
int vis[maxn];
int k;

bool isClique(){
	for(int i=1;i<=k;i++){
		for(int j=1;j<=k;j++){
			if(i!=j && g[a[i]][a[j]] == 0) return false;
		}
	}
	return true;
}

bool isMaxClique(){
	for(int i=1;i<=n;i++) vis[i] = 0;
	for(int i=1;i<=k;i++) vis[a[i]] = 1;
	for(int i=1;i<=n;i++){
		if(!vis[i]){
			for(int p=1;p<=k;p++){
				for(int q=1;q<=k;q++){
					if(p != q || (k==1)){ //这里特判k==1的情况  如样例图中查询1 8:8号结点应该输出Not Maximal 
						if(g[i][a[p]]==1 && g[i][a[q]] == 1){
							return false;
						}
					}
				}
			}
		}
	}
	return true;
}

int main(){
	scanf("%d%d",&n,&ne);
	for(int i=1;i<=ne;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		g[u][v] = g[v][u] = 1;
	}
	scanf("%d",&m);
	for(int i=1;i<=m;i++){
		scanf("%d",&k);
		for(int j=1;j<=k;j++) scanf("%d",&a[j]);
		bool flag1 = isClique(); 
		bool flag2 = isMaxClique();
		if(flag1 == false){
			puts("Not a Clique");
		}else{
			if(flag2 == false){
				puts("Not Maximal");
			}else{
				puts("Yes");
			}
		}
	}
	return 0;
} 

1143 Lowest Common Ancestor 29/30

思路：给先序序列建树，找最近公共祖先
一个点段错误，有时间再补。。

#include<bits/stdc++.h>
using namespace std;

const int maxn = 10010;
int m,n;
struct node{
	int v;
	node *l;
	node *r;
};
int a[maxn];
set<int> se;
int deep[maxn];
int fa[maxn];

void getDeep(node *root,int depth,int father){
	if(root == NULL) return;
	deep[root->v] = depth;
	fa[root->v] = father;
	if(root->l != NULL) getDeep(root->l,depth+1,root->v);
	if(root->r != NULL) getDeep(root->r,depth+1,root->v);
}

void build(int pos,node *root){
	if(root == NULL) return;
	node* fa = root;
	int v = a[pos];
	if(v < fa->v){
		if(fa->l == NULL){
			node *cur = new node();
			cur->v = v;
			cur->l = NULL;
			cur->r = NULL;
			fa->l = cur;
			return;
		}
		build(pos,fa->l);
	}else{
		if(fa->r == NULL){
			node *cur = new node();
			cur->v = v;
			cur->l = NULL;
			cur->r = NULL;
			fa->r = cur;
			return;
		}
		build(pos,fa->r);
	}
	return;
}

int main(){
	scanf("%d%d",&m,&n);
	for(int i=1;i<=n;i++) {
		scanf("%d",&a[i]);
		se.insert(a[i]);
	}
	node *Root = new node();
	Root->l = NULL;
	Root->r = NULL;
	if(n>=1) Root->v = a[1];
	else Root->v = 0; 
	if(n>=2) for(int i=2;i<=n;i++) build(i,Root);
	getDeep(Root,1,0);
	for(int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		bool visu = true;
		bool visv = true;
		if(se.find(u) == se.end()) visu = false;
		if(se.find(v) == se.end()) visv = false;
		if(visu == false || visv == false){
			if(visu==false && visv==false) printf("ERROR: %d and %d are not found.
",u,v);
			else if(visu == false) printf("ERROR: %d is not found.
",u);
			else printf("ERROR: %d is not found.
",v);
		}else{
			if(deep[u] < deep[v]){
				int p = u;
				int q = v;
				while(deep[q] > deep[p]) q = fa[q];
				if(q == p) printf("%d is an ancestor of %d.
",u,v);
				else{
					while(p != q){
						p = fa[p];
						q = fa[q];
					}
					printf("LCA of %d and %d is %d.
",u,v,p);
				}
			}else{
				int p = u;
				int q = v;
				while(deep[p] > deep[q]) p = fa[p];
				if(p == q) printf("%d is an ancestor of %d.
",v,u);
				else{
					while(p != q){
						p = fa[p];
						q = fa[q];
					}
					printf("LCA of %d and %d is %d.
",u,v,p);
				}
			}
		}
	}
	return 0;
}