Testing Round #11

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A. Up the hill

构造

#include <iostream>using namespace std;int a,b;int main() {    cin >> a >> b;    for (int i = b+1; i > 0; i--) cout << i << ' ';    for (int i = 1; i <= a; i++) cout << i+b+1 << ' ';    return 0;}

B. New York Hotel

模拟或者数学题

计算出离四个角最近的四个点，然后暴力枚举

//      whn6325689#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define eps 1e-9#define INF 0x3f3f3f3f#define LLINF 1LL<<62template<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);}//-----------------------------------const int MAXN=1e5+9;ll dis(ll x1,ll y1,ll x2,ll y2){    return abs(x1-x2)+abs(y1-y2);}int n,m,cc,hh;ll a,b;ll ans=LLINF,id=-1;ll x[4],y[4];ll milu=LLINF,mild=LLINF,miru=LLINF,mird=LLINF,maxx;int main(){read(n),read(m);read(cc);for(int i=0;i<cc;i++){read(a),read(b);if(dis(0,0,a,b) <= milu){            milu=dis(0,0,a,b);            x[0]=a,y[0]=b;        }        if(dis(n,0,a,b) <= mild){            mild=dis(n,0,a,b);            x[1]=a,y[1]=b;        }        if(dis(0,m,a,b) <= miru){            miru=dis(0,m,a,b);            x[2]=a,y[2]=b;        }        if(dis(n,m,a,b) <= mird){            mird=dis(n,m,a,b);            x[3]=a,y[3]=b;        }}read(hh);for(int i=1;i<=hh;i++){read(a),read(b);maxx=0;for(int j=0;j<4;j++)maxx=max(maxx,dis(a,b,x[j],y[j]));if(maxx<ans){ans=maxx;id=i;}}write(ans),putchar('\n'),write(id),putchar('\n');return 0;}

C. Deciphering

二分图的最优匹配，直接上KM模板

//      whn6325689#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>using namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;typedef vector<int> vi;#define CLR(x,y) memset(x,y,sizeof(x))#define mp(x,y) make_pair(x,y)#define pb(x) push_back(x)#define lowbit(x) (x&(-x))#define MID(x,y) (x+((y-x)>>1))#define eps 1e-9#define INF 0x3f3f3f3f#define LLINF 1LL<<62template<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);}//-----------------------------------const int MAXN=66;int n, m;int adj[MAXN][MAXN];int labelx[MAXN], usedx[MAXN], lnk[MAXN];int labely[MAXN], usedy[MAXN];int mat;char s1[2000005];char s2[2000005];bool path(int i){    usedx[i]=1;    for(int j=0; j<m; j++)    {        if(!usedy[j] && adj[i][j]!=-INF && !abs(adj[i][j]-labelx[i]-labely[j]))        {            usedy[j]=1;            if(lnk[j]==-1 || path(lnk[j]))            {                lnk[j]=i;                return true;            }        }    }    return false;}int match(){    mat=0;    CLR(lnk,-1);    CLR(labely,0);    for(int i=0; i<n; i++)    {        labelx[i]=0;        for(int j=0; j<m; j++)            if(adj[i][j] > labelx[i])                labelx[i] = adj[i][j];    }    for(int k=0; k<n; k++)    {        while(1)        {            CLR(usedx,0);            CLR(usedy,0);            if(path(k))            {                mat++;                break;            }            int del=INF;            for(int i=0; i<n; i++)                if(usedx[i])                    for(int j=0; j<m; j++)                        if(!usedy[j] && adj[i][j]!=-INF)                            del=min(del,labelx[i]+labely[j]-adj[i][j]);            if(del==0 || del==INF)                break;            for(int i=0; i<n; i++)                if(usedx[i])                    labelx[i] -= del;            for(int j=0; j<m; j++)                if(usedy[j])                    labely[j] += del;        }    }    int sum=0;    for(int i=0; i<n; i++)        sum+=labelx[i];    for(int j=0; j<m; j++)        sum+=labely[j];    return sum;}int main(void){    int len, k;    int a,b;    read(len),read(k);    gets(s1),gets(s2);    n=m=k;    for(int i=0; i<len; i++)    {        if(islower(s1[i]))            a=s1[i]-'a';        else            a=s1[i]-'A'+26;        if(islower(s2[i]))            b=s2[i]-'a';        else            b=s2[i]-'A'+26;        adj[a][b]++;    }    int r=match();    write(r),putchar('\n');    for(int i=0; i<k; i++)        for(int j=0; j<k; j++)            if(lnk[j]==i)                if(j>=26)                    printf("%c", j+'A'-26);                else                    printf("%c", j+'a');    puts("");    return 0;}

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