2019 ACM-ICPC 西安邀请赛 A C D L M (/签到题 题解)

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A

Tasks
上来以为DP 结果直接贪也是楞了

#include <bits/stdc++.h>
#define fastio ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
#define int long long

const int maxn = 1e5 + 5 ;

int a[maxn];

signed main() {
    fastio;
    int n, m;
    cin >> n >> m;

    for(int i = 1 ; i <= n; i++) {
        cin >> a[i];
    }

    int res = 0;
    int ans = 0;
    sort(a + 1, a + 1 + n);
    for(int i = 1; i <=n; i++) {
        if(res + a[i] <= m)
            res += a[i], ans++;
        else
            break;
    }

    cout << ans << endl;

    return 0;
}

C

** Angel’s Journey**
画图计算公式 简单计算下

#include <bits/stdc++.h>
#define fastio ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
#define int long long

const int maxn = 1e5 + 5 ;
const double pi = acos(-1.0);

double rdis(double x, double y, double a, double b) {
    return sqrt((x - a) * (x - a) + (y - b) * (y - b));
}

signed main() {
    //  fastio;
    int T;
    cin >> T;
    while(T--) {
        double rx, ry, r, x, y;

        cin >> rx >> ry >> r >> x >> y;

        if(x < rx - r || x > rx + r) {
            if(x < rx - r) {
                double ans = rdis(rx - r, ry, x, y);
                ans += pi / 2 * r;
                printf("%.4lf\n", ans);
            } else {
                double ans = rdis(rx + r, ry, x, y);
                ans += pi / 2 * r;
                printf("%.4lf\n", ans);
            }
        } else {
            double d = rdis(x, y, rx, ry);
            double qd = sqrt(d * d - r * r);

            double a1 = atan(abs(x-rx) / abs(ry-y));
            double a2 = atan(qd / r);

            double j = pi / 2 - a1 - a2;

            double xl = j * r;

         //   cout<<qd<<"    "<<a1<<"   "<<a2<<"   "<<j<<"  "<<xl<<endl;

            double ans = xl;
            ans += qd;
            ans += pi / 2 * r;
            printf("%.4lf\n", ans);
        }
    }
    return 0;
}

D

Miku and Generals
把物品有关系 一开始dfs分好 变成一个必选其中一个的
而孤立点 就是一半的01背包

#include <bits/stdc++.h>
#define fastio ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
#define int long long

const int maxn = 1e6 + 5 ;
const double pi = acos(-1.0);

int n, m;
int dp[maxn];
int a[maxn];

int head[3005], cnt;
int nxt[10000], to[10000];

struct node {
	int v, w;
} wu[205];

void ade(int a, int b) {
	to[++cnt] = b;
	nxt[cnt] = head[a];
	head[a] = cnt;
}

int s1,s2;
int vis[maxn];

void dfs(int x) {
	if(vis[x]==1) {
		if(s1==-1) s1=0;
		s1+=a[x];
	} else {
		if(s2==-1) s2=0;
		s2+=a[x];
	}

	for(int i = head[x]; i; i = nxt[i]) {
		if(vis[to[i]]) continue;
		vis[to[i]] = (vis[x] == 1 ? 2 : 1);
		dfs(to[i]);
	}
}

signed main() {
//	fastio;
	int T;
	cin >> T;
	while(T--) {
		int sum=0;
		memset(head,0,sizeof(head));
		memset(vis,0,sizeof(vis));
		memset(dp,0,sizeof(dp));
		cnt=0;
		cin >> n >> m;
		for(int i = 1; i <= n; i++) {
			cin >> a[i];
			a[i]/=50;
			sum+=a[i];
		}

		for(int i = 1; i <= m; i++) {
			int u, v;
			cin >> u >> v;
			ade(u,v),ade(v,u);
		}

		int wus=0;
	
		for(int i=1; i<=n; i++) {
			if(vis[i]) continue;
			vis[i]=1;
			s1=-1,s2=-1;
			dfs(i);
			wu[++wus]=node {s1,s2};
		}
	
		for(int i=1; i<=wus; i++) {
			for(int j=sum/2; j>=0; j--) {
				if(wu[i].w!=-1) {
					if(j>=wu[i].w&&j>=wu[i].v) {
						dp[j]=max(dp[j-wu[i].v]+wu[i].v,dp[j-wu[i].w]+wu[i].w);
					} else if(j>=wu[i].w) {
						dp[j]=dp[j-wu[i].w]+wu[i].w;
					} else if(j>=wu[i].v) {
						dp[j]=dp[j-wu[i].v]+wu[i].v;
					}
				} else {
					if(j>=wu[i].v)
						dp[j]=max(dp[j],dp[j-wu[i].v]+wu[i].v);
				}
			}
		}
		cout<<(sum-dp[sum/2])*50<<endl;
		
	}
	return 0;
}

L

Swap
直接打表上去也压力不大。。才400ms

#include <bits/stdc++.h>
#define fastio ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
#define int long long

const int maxn = 1e5 + 5 ;

set<string> S;

string s;
int len;
void dfs(int d) {
    if(d) {
        if(len % 2 == 0) {
            for(int i = 0; i < len / 2; i++) {
                swap(s[i], s[i + len / 2 ]);
            }
          //  cout << s << endl;
            if(S.find(s) != S.end())
                return ;
            S.insert(s);
            dfs(!d);
        }
        else{
            for(int i = 0; i < len / 2; i++) {
                swap(s[i], s[i + len / 2 +1]);
            }
          //  cout << s << endl;
            if(S.find(s) != S.end())
                return ;
            S.insert(s);
            dfs(!d);
        }

    } else {
        for(int i = 0; i < len - 1; i += 2) {
            swap(s[i], s[i + 1]);
        }
       // cout << s << endl;
        if(S.find(s) != S.end())
            return ;
        S.insert(s);
        dfs(!d);
    }
}

signed main() {
    fastio;
    int n=3;
   // while(n<20){
   cin>>n;
   if(n==1) cout<<1<<endl;
   else if(n==2) cout<<2<<endl;
   else{

        s="";
        for(int i = 1; i <= n; i++)
            s.push_back(i + 'a');
      //  cout << s << endl;
        len = n;
        dfs(0);
      //  cout<<n<<"     " << S.size()<<endl;
      cout<<S.size()<<endl;
        n++;
        S.clear();
   }
   // }
    return 0;
}

M

Travel
数据水了 直接贪过去的也行。。。
正解应该是二分 chk可达n点

#include <bits/stdc++.h>
#define int long long
using namespace std;
typedef long long ll;

const int maxn=1e5+10;

struct node {
	int to,dis;
};
bool vis[maxn];

vector<node>v[maxn];
int n,m,c,d,e;

bool check(int x) {
	queue<pair<int,int> > que;
	que.push(make_pair(1,x*e));
	while(!que.empty()){
		pair<int,int> p=que.front();
		que.pop(); 
		if(p.first==n) return 1; 
		if(vis[p.first]) continue ;
		vis[p.first]=1;
		for(int i=0;i<v[p.first].size();i++){
			if(p.second>=1&&v[p.first][i].dis<=x*d){
				que.push(make_pair(v[p.first][i].to,p.second-1));
			}
		}
	}
	return 0;
}

signed main() {
	cin>>n>>m>>c>>d>>e;
	memset(vis,0,sizeof(vis));
	for(int i=0; i<m; i++) {
		int x,y,z;
		cin>>x>>y>>z;
		v[x].push_back(node {y,z});
		v[y].push_back(node {x,z});
	}
	
	int l=0,r=1e5,mid;
	while(r-l>1) {
		memset(vis,0,sizeof(vis));
		mid=(l+r)/2;
		if(check(mid)) r=mid;
		else l=mid;
	}
	cout<<r*c<<endl;
	return 0;
}

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