UCF题目记录

UCF Local Programming Contest 2013

全是简单模拟题，没啥好讲的

UCF Local Programming Contest 2012

A.Wall Street Monopoly
区间dp，预处理一个前缀和，然后写一个树状数组查询区间最值就完事了。

#include
#define lowbit(x) x&(-x)
using namespace std;
typedef long long ll;
const int N = 1e2+5;

ll sum[N],Max[N],dp[N][N],y[N];

inline void update(int l){
	Max[l] = y[l];
	for(int i = 1;i < lowbit(l);i <<= 1) Max[l] = max(Max[l],Max[l-i]);
	return ;
}

inline ll query(int l,int r){
	ll res = y[r];
	while(l <= r){
		res = max(res,y[r]);
		for(--r;r-l >= lowbit(r);r -= lowbit(r)) res = max(res,Max[r]);
	}
	return res;
}

void init(int n){
	for(int i = 1;i <= n;++i){
		for(int j = 1;j <= n;++j)
			dp[i][j] = 2e18;
		dp[i][i] = 0;
	}
	return ;
}

int main(){
	int n,T,cas = 0;
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		memset(Max,0,sizeof(Max));
		memset(sum,0,sizeof(sum));
		init(n);
		for(int i = 1;i <= n;++i){
			scanf("%lld%lld",&sum[i],&y[i]);
			update(i);
			sum[i] += sum[i-1];
		}
		
		printf("The minimum cost for lot #%d is ",++cas);
		if(n == 1) printf("$0.\n\n");
		else{
			for(int i = 2;i <= n;++i){
				int pre = i-1;
				int now = i;
				dp[pre][now] = min(sum[pre]-sum[pre-1],y[pre])*min(y[now],sum[now]-sum[now-1]);
			}
			for(int l = 3;l <= n;++l){
				for(int i = 1;i+l-1 <= n;++i){
					int j = i+l-1;
					for(int k = i;k < j;++k){
						ll t1 = min(sum[k]-sum[i-1],query(i,k));
						ll t2 = min(sum[j]-sum[k],query(k+1,j));
						dp[i][j] = min(dp[i][j],dp[i][k]+dp[k+1][j]+t1*t2);
					}
				}
			}
			printf("$%lld.\n\n",dp[1][n]*100);
		}
	}
	return 0;
}

F.Metallic Equipment Rigid
数学题，判线段是否在圆内或者和圆相交。

#include 
using namespace std;
const int N = 1e3+5;
const double INFINV=1e10;
const double ESP=1e-8;

vector<pair<double,double> > V,Y;
typedef pair<double,double> PAIR;
double r[N];
bool vis[N];
vector <int> ans;

bool judge(PAIR P,PAIR yuan,double R)// 判断是否在圆内
{
    if( (P.first-yuan.first)*(P.first-yuan.first) + (P.second-yuan.second)*(P.second-yuan.second) -R*R <=0)
        return 1;
    return 0;
}
bool Judis(PAIR P1,PAIR P2,PAIR yuan,double R){
    if(judge(P1,yuan,R)&&judge(P2,yuan,R))//都在圆内 不相交
        return true;
    if(!judge(P1,yuan,R)&&judge(P2,yuan,R)||judge(P1,yuan,R)&&!judge(P2,yuan,R))//一个圆内一个圆外 相交
        return true;
    double A,B,C,dist1,dist2,angle1,angle2;//Ax+By+C=0;//(y1-y2)x +(x2-x1)y +x1y2-y1x2=0
    if(P1.first==P2.first)
        A=1,B=0,C= -P1.first;
    else if(P1.second==P2.second)
        A=0,B=1,C= -P1.second;
    else
    {
        A = P1.second-P2.second;
        B = P2.first-P1.first;
        C = P1.first*P2.second - P1.second*P2.first;
    }
    dist1 = A * yuan.first + B * yuan.second + C;
    dist1 *= dist1;
    dist2 = (A * A + B * B) * R * R;
    if (dist1 > dist2) return false;//点到直线距离大于半径r  不相交
    angle1 = (yuan.first - P1.first) * (P2.first - P1.first) + (yuan.second - P1.second) * (P2.second - P1.second);
    angle2 = (yuan.first - P2.first) * (P1.first - P2.first) + (yuan.second - P2.second) * (P1.second - P2.second);
    if (angle1 > 0 && angle2 > 0) return true;//余弦都为正，则是锐角 相交
    return false;//不相交
}
 
int main()
{
    int n,m,T,cas = 0;
    double x,y,rr;
    scanf("%d",&T);
    while(T--){
        V.clear();
        Y.clear();
        ans.clear();
        scanf("%d%d",&n,&m);
        for(int i = 1;i <= n;++i){
        	vis[i] = 0;
        	scanf("%lf%lf%lf",&x,&y,&r[i]);
        	Y.push_back(make_pair(x,y));
		}
        for(int i = 1;i <= m;++i){
            scanf("%lf%lf",&x,&y);
            V.push_back(make_pair(x,y));
        }
        
        for(int i = 1;i < V.size();++i){
        	for(int j = 0;j < Y.size();++j){
        		if(vis[j+1]) continue;
        		if(Judis(V[i-1],V[i],Y[j],r[j+1])){
        			vis[j+1] = 1;
        			ans.push_back(j+1);
				}
			}
		}
		printf("Compound #%d: ",++cas);
		if(ans.size()){
			sort(ans.begin(),ans.end());
			printf("Reptile triggered these cameras:");
			for(int i = 0;i < ans.size();++i) printf(" %d",ans[i]);
			printf("\n\n");
		}
		else printf("Rigid Reptile was undetected\n\n");
    }
    return 0;
}

UCF Local Programming Contest 2015

F.Balanced Strings
dp就完事了，因为所有子串都是和谐的串，所以两种音应该交叉进行，注意第一个字符的处理即可。

#include 
using namespace std;
typedef long long ll;
const int N = 1e2+5;
const ll INF = 2e18;
const double eps = 1e-6;
const double Pi = 3.14159265358979;

ll dp[105][3];
map<char,int> mp;

int main() {
	int T;
    scanf("%d",&T);
    
    string s1 = "aeiouy";
    for(int i=0;i<6;i++) mp[s1[i]] = 1;
	int ct = 1;
    while(T--) {
		memset(dp,0,sizeof(dp));
		string s;
		cin>>s;
		printf("String #%d: ",ct++);
		
		if(s[0]=='?') dp[0][1] = 6,dp[0][2] = 20;
		else if(mp[s[0]]) dp[0][1] = 1;
		else dp[0][2] = 1;
		
		for(int i=1;i<s.length();i++){
			if(s[i]!='?'){
				if(mp[s[i]]) dp[i][1] = dp[i-1][2];
				else dp[i][2] = dp[i-1][1];
			}
			else dp[i][2] = dp[i-1][1]*20,dp[i][1] = dp[i-1][2]*6;
		}
		int len = s.length();
		printf("%lld\n\n",dp[len-1][1]+dp[len-1][2]);
    }
    return 0;
}

H.Reach for the Stars
题解是状压。。。现场写了个搜索。。。应该是数据水了。。。

#include 
using namespace std;
typedef long long ll;
const int N = 1e2+5;
const ll INF = 2e18;
const double eps = 1e-6;
const double Pi = 3.14159265358979;
 
char mp[15][15];
int vis[15][15],ans,steps;
struct node{
    int x, y;
    int p;
    node(){}
    node(int x,int y,int p):x(x),y(y),p(p){}
    bool operator < (node c){
        return p < c.p;
    }
} s[100];

inline bool check1(int x,int y){
	if(mp[x-1][y] != '#' || mp[x+1][y] != '#' || mp[x][y-1] != '#' || mp[x][y+1] != '#') return 0;
	return 1;
}

inline bool check2(int x,int y){
	if(!vis[x-1][y] || !vis[x+1][y] || !vis[x][y-1] || !vis[x][y+1]) return 1;
	return 0;
}

inline bool judge(int n,int m){
	for(int i = 1;i <= n;++i)
		for(int j = 1;j <= m;++j)
			if(!vis[i][j] && mp[i][j] == '#') return 0;
	return 1;
}

inline void update(const int &i,const int &j,int &rest){
    ++vis[i][j];
    if(vis[i][j] == 1) rest -= 1;
    return ;
}

inline void Update(const int &i,const int &j,int &rest) {
    update(i,j,rest);
    update(i+1,j,rest);
    update(i,j+1,rest);
    update(i-1,j,rest);
    update(i,j-1,rest);
    return ;
}

inline void hs(const int &i,const int &j,int &rest) {
    --vis[i][j];
    if(vis[i][j] == 0) rest += 1;
    return ;
}
 
inline void Hs(const int &i,const int &j,int &rest) {
    hs(i,j,rest);
    hs(i+1,j,rest);
    hs(i,j+1,rest);
    hs(i-1,j,rest);
    hs(i,j-1,rest);
    return ;
}

void dfs(int now,int res,int rest) {
    if(rest == 0) {
        ans = min(ans,res);
        return;
    }
    if(res+(rest+4)/5 >= ans) return;
 
    int x = s[now].x, y = s[now].y;
    if(now == steps){
        ++res;
        Update(x,y,rest);
        if(rest == 0) ans = min(ans,res);
        Hs(x,y,rest);
        --res;
        return;
    }
    
	if(check2(x,y)){
		++res;
	    Update(x,y,rest);
	    dfs(now+1,res,rest);
	    Hs(x,y,rest);
	    --res;
	}
    dfs(now+1,res,rest);
    return;
}

int main(){
    int n,m,T,cas = 0;
    scanf("%d",&T);
    while(T--){
    	memset(mp,0,sizeof(mp));
        memset(vis,0,sizeof(vis));
        int sum = 0;
		scanf("%d%d",&n,&m);
        printf("Image #%d: ",++cas);
		for(int i = 1;i <= n;++i){
        	scanf("%s",mp[i]+1);
        	for(int j = 1;j <= m;++j)
        		if(mp[i][j] == '#') sum++;
		}
 
        ans = 0;steps = 0;
        for(int i = 2;i < n;++i)
            for(int j = 2;j < m;++j)
                if(mp[i][j] == '#' && check1(i,j)){
                	++ans;
					vis[i][j] = vis[i - 1][j] = vis[i][j - 1] = vis[i + 1][j] = vis[i][j + 1] = 1;
   					s[++steps] = node(i,j,min(min(i-1,j-1),min(n-i,m-j)));
				}

        if(!judge(n,m)) printf("impossible\n\n");
        else{
            memset(vis, 0, sizeof(vis));
			int res = 0;
            for(int i = 1;i <= m;++i){
                if(mp[1][i] == '#' && vis[1][i] == 0) ++res,Update(2,i,sum);
                if(mp[n][i] == '#' && vis[n][i] == 0) ++res,Update(n-1,i,sum);
            }
            for(int i = 1;i <= n;++i){
                if(mp[i][1] == '#' && vis[i][1] == 0) ++res,Update(i,2,sum);
                if(mp[i][m] == '#' && vis[i][m] == 0) ++res,Update(i,m-1,sum);
            }
            sort(s+1,s+1+steps);
            dfs(1,res,sum);
            printf("%d\n\n", ans);
        }
    }
    return 0;
}

I.Longest Path
竞赛图，哈密顿路径，自行百度即可。

#include
using namespace std;
typedef long long ll;
const int N = 5e2+5;
const int MAXM = 200100;

int mp[N][N];

int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        memset(mp,0,sizeof(mp));
        scanf("%d",&n);
        for(int i = 0;i < n;++i)
        	for(int j = 0;j < n;++j)
        		scanf("%d",&mp[i][j]);
        
        vector<int> v;
        v.push_back(0);
        for(int i = 1; i < n; i++){
            int tmp = v.size()-1;
            while(tmp >= 0 && !mp[v[tmp]][i]) tmp--;
            v.insert(v.begin()+tmp+1,i);
        }
        for(int i = 0; i < n; i++) printf("%d%c",v[i]+1,i == n-1 ?'\n' :' ');
    }
    return 0;
}

J.You Shall Pass
最小割，和小M的作物一样的题目，建个图跑Dinic就完事了

#include 
typedef long long ll;
using namespace std;
const int N = 4e4+5;
const int M = 1e6+5;
const int INF = 0x3f3f3f3f;

struct node{
	int to,next,flow;
	node(){}
	node(int to,int next,int flow):to(to),next(next),flow(flow){}
} e[M << 1];
int S,T,tot,head[N],q[N],dep[N],cur[N];

void init(int n){
	memset(head,-1,sizeof(head));
	tot = -1;
	S = 0;
	T = n+n*n*2+1;
	return ;
}

void add(int u,int v,int w){
	e[++tot] = node(v,head[u],w);
	head[u] = tot;
	e[++tot] = node(u,head[v],0);
	head[v] = tot;
	return ;
}

inline bool bfs(){
	memset(dep,0,sizeof(dep));
	int st = 0,ed = 1;
	q[st] = S;dep[S] = 1;
	
	while(st != ed){
		int u = q[st++];
		for(int i = head[u];~i;i = e[i].next){
			int v = e[i].to;
			if(!dep[v] && e[i].flow > 0){
				dep[v] = dep[u]+1;
				if(v == T) return 1;
				q[ed++] = v;
			}
		}
	}
	return dep[T];
}

inline int dfs(int u,int nowflow){
	if(u == T || !nowflow) return nowflow;
	int totflow = 0;
	for(int &i = cur[u];~i;i = e[i].next){
		int v = e[i].to;
		if(dep[v] == dep[u]+1 && e[i].flow > 0 && nowflow-totflow > 0){
			int tflow = dfs(v,min(nowflow-totflow,e[i].flow));
			e[i].flow -= tflow;
			e[(i^1)].flow += tflow;
			totflow += tflow;
			nowflow -= tflow;
			if(nowflow <= 0) break;
		}
	}
	return totflow;
}

inline int Dinic(){
	int ans = 0;
	while(bfs()) memcpy(cur,head,sizeof(head)),ans += dfs(S,INF);
	return ans;
}

int main(){
    int n,TT;
	scanf("%d",&TT);
	while(TT--){
		scanf("%d",&n);
		init(n);
		double t;
		
		int ans = 0;
		for(int i = 1;i <= n;++i){
			scanf("%lf",&t);
			int tt = round(t*100);
			add(S,i,tt);
			ans += tt;
		}
		for(int i = 1;i <= n;++i){
			scanf("%lf",&t);
			int tt = round(t*100);
			add(i,T,tt);
			ans += tt;
		}
		for(int i = 1;i <= n;++i){
			for(int j = 1;j <= n;++j){
				scanf("%lf",&t);
				int tt = round(t*100);
				add(n+(i-1)*n+j,i,INF);
				add(n+(i-1)*n+j,j,INF);
				add(S,n+(i-1)*n+j,tt);
				add(i,n+n*n+(i-1)*n+j,INF);
				add(j,n+n*n+(i-1)*n+j,INF);
				add(n+n*n+(i-1)*n+j,T,tt);
				ans += (tt << 1);
			}
		}
		printf("%.2f\n",(double)(ans-Dinic())/100.0);
	}
    return 0;
}