codeforces #270 ABCD

Codeforces Round #270

A - Design Tutorial: Learn from Math

题意：给出n，求出两个合数x和y使x+y=n。

题解：暴力筛合数，然后暴力找

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll __int64
 #define usint unsigned int
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(int i=0;i<(n);i++)
 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1.out","w",stdout)

 bool a[];

 int main() {
     int x,i,n,j;
     RD(n);
     int n2=n/;
     for(x=; x<=n2; x++) {
         if(a[x]==) {
             for(i=x+x; i<=n; i+=x) {
                 a[i]=;
             }
         }
     }
     for(x=;x<=n2;x++){
         if(a[x]== && a[n-x]==)break;
     }
     printf("%d %d\n",x,n-x);
     return ;
 }

B - Design Tutorial: Learn from Life

题意：一堆人从1楼坐电梯，电梯一次最多装K人，给出各个人要去的各个楼层，上下电梯时间忽略不计，电梯移动一层需要1秒，求最少需要多少时间才能送完所有人并且电梯回到1楼。

题解：贪心，优先送最高的，空位顺便送低的。

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll __int64
 #define usint unsigned int
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(int i=0;i<(n);i++)
 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1.out","w",stdout)

 int a[];

 int main() {
     int n,k,i,j,x,ans;
     RD2(n,k);
     REP(i,n) {
         RD(x);
         a[x]++;
     }
     x=;
     ans=;
     for(i=;i>=;i--) {
         if(a[i]>) {
             a[i]-=x;
             int t=a[i]/k;
             int tt=t*k;
             int ttt=tt>=a[i]?t:t+;
             ans+=ttt*(i-)*;
             x=ttt*k-a[i];
             //printf("%d %d %d\n",i,ttt,ans);
         }
     }
     WN(ans);
     return ;
 }

C - Design Tutorial: Make It Nondeterministic

题意：每个人可以用自己的姓或者名字作为ID，给出各个人的名字，给出一个排名序列，若有一种选择ID的方法可以使排名正确，则输出YES，否则输出NO。

题解：贪心，排名1的人选字典序少的那个，然后都优先选字典序小的，不行了就选大的，再不行就NO。

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll __int64
 #define usint unsigned int
 #define mz(array) memset(array, 0, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(int i=0;i<(n);i++)
 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1.out","w",stdout)

 char s[][][];

 int cmp(char a[],char b[]) {
     int la=strlen(a),lb=strlen(b);
     int l=min(la,lb);
     int i;
     for(i=; i<l; i++) {
         if(a[i]>b[i])return ;
         if(b[i]>a[i])return -;
     }
     if(la>lb)return ;
     if(lb>la)return -;
     return ;
 }

 int main() {
     int n;
     RD(n);
     REP(i,n) {
         scanf(" %s%s",s[i][],s[i][]);
         //printf("%d %s!%s!\n",i,s[i][0],s[i][1]);
     }
     char q[];
     int x,y=,z,j;
     RD(x);
     y=x-;
     if(cmp(s[y][],s[y][])==-) z=;
     else z=;
     bool flag=;
     REP(i,n-) {
         RD(x);
         x--;
         if(cmp(s[x][],s[x][])==-) j=;
         else j=;
         if(cmp(s[x][j] , s[y][z])==-){
             if(cmp(s[x][j^], s[y][z])==-){
                 flag=;
                 break;
             }else{
                 y=x;
                 z=j^;
             }
         }else{
             y=x;
             z=j;
         }
     }
     if(flag)puts("YES");
     else puts("NO");
 return ;
 }

D - Design Tutorial: Inverse the Problem

题意：给出各个点距离的邻接矩阵，判断这是不是一棵树。

题解：先根据样例特判一些情况，然后最小生成树，我是用可撸斯卡尔，当要加入的边的两个点x和y已经在同一集合，则判它们的距离是否正确。我的方法是找到一个中间点v，使x-v和v-y有路，找到(xv距离) + (vy距离)最小的那个值，当作x到y的距离。这是因为可撸斯卡尔是从小往大加边，肯定有至少一个中间点能使得这个有路。复杂度O(n^3)，略高，姿势不对就要TLE……更好的解法可以看这里：http://www.cnblogs.com/forgot93/p/4000850.html

 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include<cstdio>
 #include<cmath>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<map>
 #include<set>
 #include<stack>
 #include<queue>
 using namespace std;
 #define ll __int64
 #define usint unsigned int
 #define mz(array) memset(array, 0, sizeof(array))
 #define mf1(array) memset(array, -1, sizeof(array))
 #define minf(array) memset(array, 0x3f, sizeof(array))
 #define REP(i,n) for(int i=0;i<(n);i++)
 #define FOR(i,x,n) for(int i=(x);i<=(n);i++)
 #define RD(x) scanf("%d",&x)
 #define RD2(x,y) scanf("%d%d",&x,&y)
 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
 #define WN(x) printf("%d\n",x);
 #define RE  freopen("D.in","r",stdin)
 #define WE  freopen("1.out","w",stdout)
 const int inf=*1e9+;
 const int maxn=;
 const int maxm=maxn*maxn;

 int a[][];
 int n;

 int bx[maxm],by[maxm],bz[maxm];
 int bn;
 inline void add(const int &x,const int &y,const int &z){
     bx[bn]=x;
     by[bn]=y;
     bz[bn]=z;
     bn++;
 }

 int d[maxm];
 bool cmp(const int &x,const int &y){
     return bz[x]<bz[y];
 }

 int f[maxm];
 int gf(const int &x){
     if(f[x]==-)return x;
     f[x]=gf(f[x]);
     return f[x];
 }

 struct Edge{
     int next,v,l;
 }e[maxm];
 int head[maxn],en;

 inline void addedge(const int &x,const int &y,const int &z){
     e[en].v=y;
     e[en].l=z;
     e[en].next=head[x];
     head[x]=en;
     en++;
 }

 int dis[maxn][maxn];

 inline void gank(const int &x,const int &y){
     int mi=inf;
     for(int i=head[x];i!=-;i=e[i].next){
         if(dis[y][e[i].v]!=-){
             //printf("(%d) %d + %d = %d\n",e[i].v,e[i].l,dis[y][e[i].v],e[i].l + dis[y][e[i].v]);
             mi=min(mi , e[i].l + dis[y][e[i].v]);
         }
     }
     dis[y][x]=dis[x][y]=mi;
 }

 bool farm(){
     int i,j,k,x,y,z;
     bn=;
     REP(i,n){
         REP(j,n){
             if(i!=j){
                 if(a[i][j]==)return ;
             }else{
                 if(a[i][j]!=)return ;
             }
         }
     }
     REP(i,n){
         REP(j,i){
             if(a[i][j]!=a[j][i])return ;
             add(i,j,a[i][j]);
         }
     }
     REP(i,bn){
         d[i]=i;
     }
     sort(d,d+bn,cmp);
     mf1(f);
     //mf1(head);
     REP(i,n)REP(j,n)dis[i][j]=-;
     en=;
     mf1(head);
     REP(i,n)dis[i][i]=;
     //en=0;
     REP(i,bn){
         int q=d[i];
         int &x=bx[q],&y=by[q],&z=bz[q];
         int fx=gf(x),fy=gf(y);
         if(fx!=fy){
             addedge(x,y,z);
             addedge(y,x,z);
             dis[x][y]=z;
             dis[y][x]=z;
             f[fx]=fy;
         }else{
             if(dis[x][y]==-) gank(x,y);
             //printf("dis[%d][%d]=%d , z=%d\n",x,y,dis[x][y],z);
             if(dis[x][y]!=z)return ;
         }
     }
     return ;
 }

 int main() {
     int i;
     RD(n);
     REP(i,n)REP(j,n) RD(a[i][j]);
     if(farm())puts("YES");
     else puts("NO");
     return ;
 }