BOJ 2014新生暑假个人排位赛08 整合

A. 游戏

只要枚举边上的角度即可, 使用点乘可以轻松判断

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cctype>#include <cmath>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <algorithm>#include <climits> #define MAXN 100005#define eps 1e-5#define MOD 1000000009 #define test #define For(i,m,n) for(int i=(m);i<(n);i++)#define vecfor(iter,a) for(vector<int>::iterator iter=a.begin();iter!=a.end();iter++)#define rep(i,m,n) for(int i=(m);i<=(n);i++)#define LL long long /*author birdstorm*/using namespace std;const double pi=acos(-1.0);double a[1111][3]; double cross(double x1, double y1, double x2, double y2){    return x1*x2+y1*y2;}  int main() {     //freopen("input.txt","r",stdin);     int n, m;     int cnt=0;     while(~scanf("%d",&n)){        For(i,0,n) scanf("%lf%lf",&a[i][0], &a[i][1]);        a[n][0]=a[0][0], a[n][1]=a[0][1];        a[n+1][0]=a[1][0], a[n+1][1]=a[1][1];        a[n+2][0]=a[2][0], a[n+2][1]=a[2][1];        int cnt=n;        For(i,0,n){            double x1=a[i][0]-a[i+1][0];            double y1=a[i][1]-a[i+1][1];            double x2=a[i+1][0]-a[i+2][0];            double y2=a[i+1][1]-a[i+2][1];            double x3=a[i+2][0]-a[i+3][0];            double y3=a[i+2][1]-a[i+3][1];            if(cross(x1,y1,x2,y2)<=eps&&cross(x2,y2,x3,y3)<=eps) cnt--;        }        printf("%d\n",cnt);     }     return 0; }

B. 小妹妹送快递

题意要求的是s-t路径上最大边权的最小值

将边从小到大排序后, 依次连边建图. 

最后直接并查集查找即可

要注意即使路径上最小边权为0, 也必须使用一个小妹妹送快递

所以需要特判.

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cctype>#include <cmath>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <algorithm>#include <climits>  #define MAXN 100005#define eps 1e-5#define MOD 1000000009#define INF 1000000009  #define test  #define For(i,m,n) for(int i=(m);i<(n);i++)#define vecfor(iter,a) for(vector<int>::iterator iter=a.begin();iter!=a.end();iter++)#define rep(i,m,n) for(int i=(m);i<=(n);i++)#define LL long long  /*author birdstorm*/using namespace std;const double pi=acos(-1.0);  template<class T>inline bool read(T &n){    T x=0, tmp=1; char c=getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c=getchar();    if(c == EOF) return false;    if(c == '-') c=getchar(), tmp=-1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c=getchar();    n=x*tmp;    return true;}  template <class T>inline void write(T n) {    if(n < 0) {        putchar('-');        n=-n;    }    int len=0,data[20];    while(n) {        data[len++]=n%10;        n /= 10;    }    if(!len) data[len++]=0;    while(len--) putchar(data[len]+48);}    pair<pair<int,int>,int> edge[200005];LL a[MAXN];int vis[MAXN];struct Disjoint{    vector<int> father,rank;    Disjoint(int n):father(n),rank(n){        For(i,0,n) father[i]=i;    }    int find(int v){        return father[v]=father[v]==v?v:find(father[v]);    }    void merge(int x,int y){        int a=find(x), b=find(y);        if(a==b) return;        if(rank[a]<rank[b]){            father[a]=b;        }        else{            father[b]=a;            if(rank[b]==rank[a]) ++rank[a];        }    }    void clear(int n){        For(i,0,n) father[i]=i, rank[i]=0;    }};  bool cmp(pair<pair<int,int>,int> a, pair<pair<int,int>,int> b){    return a.second<b.second;}  int main(){    //freopen("input.txt","r",stdin);    //freopen("output.txt","w",stdout);    int u, v, w;    Disjoint D(MAXN);    int n, m;    int t;    scanf("%d",&t);    while(t--){        read(n),read(m);        int ans=0;        int tot=0;        D.clear(n+1);        For(i,0,m){            read(u),read(v),read(w);            edge[tot++]=make_pair(make_pair(u,v),w);        }        sort(edge,edge+tot,cmp);        //For(i,0,tot) printf("%d %d %d--\n",edge[i].first.first,edge[i].first.second,edge[i].second);        bool found=false;        For(i,0,tot){            D.merge(edge[i].first.first,edge[i].first.second);            if(D.find(1)==D.find(n)){                printf("%d\n",max(1,edge[i].second));                //printf("%d\n",i);                found=true;                break;            }        }        if(!found) printf("shimatta!\n");    }    return 0;}

C. 学姐点名

签到题, 有代码长度限制.....所以姿势比较难看.(其实没必要

#include <cstdio>int main(){long long n, a, sum; while(~scanf("%lld",&n)) { sum=n*(n+1)/2; for(int i=0; i<n-1; i++) scanf("%lld",&a), sum-=a; printf("%lld\n",sum);} return 0;}

D. 解码锦标赛

模拟, 下标太乱了导致当时没调出来...

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cctype>#include <cmath>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <algorithm>#include <climits> #define MAXN 100005#define eps 1e-5#define MOD 1000000009 #define test #define For(i,m,n) for(int i=(m);i<(n);i++)#define vecfor(iter,a) for(vector<int>::iterator iter=a.begin();iter!=a.end();iter++)#define rep(i,m,n) for(int i=(m);i<=(n);i++)#define LL long long /*author birdstorm*/using namespace std;const double pi=acos(-1.0);double a[3000][3000];double dp[3000][3000];  int main() {     //freopen("input.txt","r",stdin);     int n, m;     int f[10]; f[0]=1;     For(i,1,10) f[i]=f[i-1]*2;     while(scanf("%d",&n),n){        int cnt=0;        For(i,0,f[n]) dp[i][0]=1;        For(i,0,f[n]) For(j,0,f[n]) scanf("%lf",&a[i][j]);        For(i,1,n+1){            For(j,0,f[n]){                dp[j][i]=dp[j][i-1];                double ans=0;                int st=(j/f[i-1])^1;                st=st*f[i-1];                int en=st+f[i-1];                //printf("%d %d %d\n",st,en-1,j);                For(k,st,en){                    //if(k==j) continue;                    ans+=dp[k][i-1]*a[j][k];                    //printf("%d\n",k);                }                dp[j][i]*=ans;            }        }        int ret=0;        double maxx=0;        For(i,0,f[n]){            //printf("%lf ",dp[i][n]);            if(maxx<dp[i][n]){                maxx=dp[i][n];                ret=i+1;            }        }        printf("%d\n",ret);     }     return 0; }

E. 田田的算数题

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