Codeforces Round #485 (Div. 2)重點：D - Fair多源最短路

鏈接
A - Infinity Gauntlet
map

int n;
int a[N];
int main(){
    map<string,int>mp;
    string a[8];
    mp["purple"]  = 1;
    a[1] = "Power";
    mp["green"] = 2;
    a[2] = "Time";
    mp["blue"] = 3;
    a[3] = "Space";
    mp["orange"] = 4;
    a[4] = "Soul";
    mp["red"] = 5;
    a[5] = "Reality";
    mp["yellow"] = 6;
    a[6] = "Mind";
    string b;
    while(~scanf("%d" , &n)){
            int vis[10];
            memset(vis,0,sizeof vis);
        for(int i = 1; i <= n ; i ++){
            cin >> b;
            vis[mp[b]] = 1;
        }
        vector<string>v;int m = 0;
        for(int i = 1; i <= 6; i ++)
            if(!vis[i])
                m ++ , v.pb(a[i]);
        cout << m <<"\n";
        for(auto x : v)cout<<x<<"\n";
    }
}

B - High School: Become Human
好像是直接判斷就可，校內大神說需要取fabs判斷更準確附上吧

int main(){
    int a,b;
    while(~scanf("%d %d" , &a , &b)){
        double x = a*1.0,y = b*1.0;
        if(y * log(x) < x * log(y)){
            puts("<");
        }
        else if(y * log(x) > x * log(y)){
            puts(">");
        }
        else{
            puts("=");
        }
    }
}

int n,m,t,q;
const double eps=1e-10;
int main(){
    int x,y;
    double a,b;
    scanf("%d%d",&x,&y);
    a=log(x)*y;
    b=log(y)*x;
    if(fabs(a-b)<=eps)printf("=\n");
    else if(a-b<0)printf("<\n");
    else printf(">\n");

    return 0;
}

C - Three displays
枚舉中間點

int n ,a[N],b[N];
int main(){

    while(~scanf("%d", &n)){
        for(int i = 1; i <= n ; i ++){
            scanf("%d" ,&a[i]);
        }
        LL mini = inf;
        for(int i =1; i <= n ; i ++){
            scanf("%d" ,&b[i]);
        }
        for(int i = 1; i <= n ; i ++){
            int mini1 = 0x3f3f3f3f;
            int mini2 = 0x3f3f3f3f;
            for(int j = i-1; j >= 1; j --){
                if(a[j] < a[i]){
                    mini1 = min(b[j] , mini1);
                }
            }
            for(int j = i + 1; j <= n ; j ++){
                if(a[j] > a[i]){
                    mini2 = min(b[j] , mini2);
                }
            }
            if(mini1 == 0x3f3f3f3f || mini2 == 0x3f3f3f3f);
            else{
                mini = min(b[i] + (LL)mini1 + mini2 , mini);
            }
        }
        if(mini == inf)puts("-1");
        else{
            cout << mini << endl;
        }
    }
}

D - Fair
多源最短路問題，主要是要想到每種商品到各個節點的最近距離，維護一下就ok

vector<int> G[maxn];
int d[maxn][105];
int n, m, k, s;
int a[maxn];
int vis[maxn];
int num[maxn];
struct node{
	int v;
	int cnt;
};

int bfs(int cur){
	memset(vis, -1, sizeof(vis));
	queue<int> q;
	for(int i=1; i<=n; i++){
		if(a[i] == cur){
			q.push(i);
			vis[i] = 0;
		}
	}
	while(!q.empty()){
		int p = q.front();
		q.pop();
		for(int i=0; i<G[p].size(); i++){
			int temp = G[p][i];
			if(vis[temp] == -1){
				vis[temp] = vis[p]+1;
				q.push(temp);
			}

		}
	}
	for(int i=1; i<=n; i++){
		d[i][cur] = vis[i];
	}
}

int main(){
	memset(d, INF, sizeof(d));
	scanf("%d%d%d%d", &n, &m, &k, &s);
	for(int i=1; i<=n; i++){
		scanf("%d", &a[i]);
		d[i][a[i]] = 0;
	}
	int v, u;
	for(int i=1; i<=m; i++){
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for(int i=1; i<=k; i++){
		bfs(i);
	}
	for(int i=1; i<=n; i++){
		sort(d[i]+1, d[i]+k+1);
		int ans = 0;
		for(int j=1; j<=s; j++){
			ans += d[i][j];
		}
		printf("%d ",ans);
	}
}

E - Petr and Permutations
模擬水題

int n ;
int a[N];
int pos[N];
int main(){
    while(~scanf("%d" , &n)){
        for(int i = 1; i <= n ; i ++){
            scanf("%d" , &a[i]);
            pos[a[i]] = i;
        }
        int ans = 0;
        for(int i = 1; i <= n ; i ++){
            if(pos[i] == i)continue;
            int x = pos[i];
            swap(pos[i],pos[a[i]]);
            swap(a[x],a[i]);
            ans ++;
        }
        if((3*n - ans) % 2 == 0){
            puts("Petr");
        }
        else{
            puts("Um_nik");
        }
    }
    return 0;
}