2020.10.10天梯赛补题

7-9 名人堂与代金券 (25分)

　　题目链接：https://pintia.cn/problem-sets/1314102638464851968/problems/1314104356594376704

　　思路：排序输出就好。。。

#include <bits/stdc++.h>

using namespace std;
const int N = 10010;

struct node{
	string id;
	int s;
};
node a[N];
bool cmp(node a, node b){
	if(a.s == b.s) return a.id < b.id;
	return a.s > b.s;
}
int main()
{
	int n, g, k;
	scanf("%d %d %d", &n,&g, &k);
	int res = 0;
	for(int i = 1;i <= n; ++ i){
		cin >> a[i].id >> a[i].s;
		if(a[i].s >= g){
			res += 50;
		}
		else if(a[i].s < 60);
		else res += 20;	
	}
	
	sort(a + 1, a + 1 + n, cmp);
	
	cout << res << endl;
	
	int i = 1;
	while(i <= k){
		int flag = i;
		for(int j = flag;j <= n; ++ j){
			if(a[j].s != a[i].s) break;
			cout << i <<" "<< a[j].id << " " << a[j].s << endl;
			flag ++;
		}
		i = flag;
	}
}

　　7-10 链表去重 (25分)

　　https://pintia.cn/problem-sets/1314102638464851968/problems/1314104356594376705

　　思路：把链表用数组存起来，按题意输出。比赛的时候忘记补零。。。

#include <bits/stdc++.h>

using namespace std;
int n;
const int N = 100010;

struct node{
    int pos, data, next;
};
node a[N];

bool c[N];

void print(vector<node>v1)
{
    if(v1.size() == 1){
        printf("%05d %d %d\n",v1[0].pos,v1[0].data,-1);    
        return;
    }
    printf("%05d %d %05d\n",v1[0].pos,v1[0].data,v1[1].pos);
    
    for(int i = 1;i < v1.size();i ++)
    {
        if(i != v1.size() - 1){
            printf("%05d %d %05d\n",v1[i].pos,v1[i].data,v1[i + 1].pos);
        }
        else{
            printf("%05d %d %d\n",v1[i].pos,v1[i].data,-1);
        }
    }
}

vector<node> ans1, ans2;
int main()
{    
//    ios::sync_with_stdio(false);    
    int head;
    scanf("%d %d", &head, &n);
    for(int i = 0;i < n; ++ i){
        int x, y, z;
         scanf("%d %d %d",&x, &y, &z);
         a[x] = {x, y, z};
    }    
    
    int h = head;
    
    for(int i = head; i != -1; i = a[i].next){
    
        int val = abs(a[i].data);
        if(c[val]){
            ans2.push_back(a[i]);
        }
    
        else{
            ans1.push_back(a[i]);
            c[val] = true;
        }
    }
//    if(ans1.size())
//    print(ans1);
//    if(ans2.size())
//    print(ans2);
    
    
    for(int i = 0;i < ans1.size(); ++ i){
        printf("%05d %d ", ans1[i].pos, ans1[i].data);
        if(i == ans1.size() - 1){
            cout << "-1\n";
        }
        else printf("%05d\n",ans1[i+1].pos);
    }

    for(int i = 0;i < ans2.size(); ++ i){
        printf("%05d %d ", ans2[i].pos, ans2[i].data);
        if(i == ans2.size() - 1){
            cout << "-1\n";
        }
        else printf("%05d\n", ans2[i+1].pos);
    }

}

7-11 部落 (25分)

　https://pintia.cn/problem-sets/1314102638464851968/problems/1314104458302054400

思路：并查集，可以高效的查看这俩是不是一个圈子的...

#include <bits/stdc++.h>

using namespace std;
const int N = 10000;
int n, m, cnt,c[N] , pre[N], k[N][N], num;
bool a[N], b[N];
set<int> st;
int find(int x){
	int r = x;
	while(pre[r] != r){
		r = pre[r];
	}
	return r;
}

void Union(int a, int b){
	int x = find(a), y = find(b);
	pre[x] = y;
}

int main()
{
	for(int i = 1;i <= 10000; ++ i) pre[i] = i;
	
	cin >> n;
	for(int i = 1;i <= n; ++ i){
		int m;
		scanf("%d", &m);
		c[i] = m;
		for(int j = 1;j <= m; ++ j){
			int x;
			scanf("%d",&x);
			k[i][j] = x;
			st.insert(x);
		}
		
		for(int j = 1;j <= m - 1; ++ j)
			Union(k[i][j], k[i][j+1]);
				
	}
	for(int i = 1;i <= 10000; ++ i) pre[i] = find(i);
	
	set<int>::iterator it = st.begin();
	for(;it != st.end(); ++ it){
		if(!a[pre[*it]]) cnt ++, a[pre[*it]] = true;
	}
	
	cout <<st.size()<<" "<< cnt << endl;
	int q;
	cin >> q;
	while(q--){
		int x, y;
		cin >> x >> y;
		if(pre[x] == pre[y]) puts("Y");
		else puts("N");
	}
}

　　7-12 月饼 (25分)

　　https://pintia.cn/problem-sets/1314102638464851968/problems/1314104458302054401

　　思路：贪心，优先选择单位价格大的月饼

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 1010;

struct node{
	double ton, price;
}a[N];

bool cmp(node a, node b){
	return ((double)a.price / (double)a.ton) > ((double)b.price / (double)b.ton);
}

int main()
{
	int n, v;
	cin >> n >> v;
	for(int i = 1;i <= n; ++ i){
		double x;
		cin >> x;
		a[i].ton = x;
//		a[i].price = y;
	}
	for(int i = 1;i <= n; ++ i){
		double x;
		cin >> x;
		a[i].price = x;
	}	
	sort(a + 1, a + 1 + n, cmp);

	double ans = 0;
	int tot = 1;
	while(v && tot <= n){
		if(a[tot].ton <= v){
			ans += (double)a[tot].price;
			v -= a[tot].ton;
		}
		else{
			ans += (double) v * (double)a[tot].price / (double)a[tot].ton;
			v = 0;
		}
		tot ++;
	}
	printf("%.2lf\n", ans);
}