标签：des style blog http color io os ar for
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
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define ll __int64 14 #define usint unsigned int 15 #define mz(array) memset(array, 0, sizeof(array)) 16 #define minf(array) memset(array, 0x3f, sizeof(array)) 17 #define REP(i,n) for(int i=0;i<(n);i++) 18 #define FOR(i,x,n) for(int i=(x);i<=(n);i++) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("D.in","r",stdin) 24 #define WE freopen("1.out","w",stdout) 25 26 bool a[1111111]; 27 28 29 int main() { 30 int x,i,n,j; 31 RD(n); 32 int n2=n/2; 33 for(x=2; x<=n2; x++) { 34 if(a[x]==0) { 35 for(i=x+x; i<=n; i+=x) { 36 a[i]=1; 37 } 38 } 39 } 40 for(x=4;x<=n2;x++){ 41 if(a[x]==1 && a[n-x]==1)break; 42 } 43 printf("%d %d\n",x,n-x); 44 return 0; 45 }
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。
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define ll __int64 14 #define usint unsigned int 15 #define mz(array) memset(array, 0, sizeof(array)) 16 #define minf(array) memset(array, 0x3f, sizeof(array)) 17 #define REP(i,n) for(int i=0;i<(n);i++) 18 #define FOR(i,x,n) for(int i=(x);i<=(n);i++) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("D.in","r",stdin) 24 #define WE freopen("1.out","w",stdout) 25 26 int a[2222]; 27 28 int main() { 29 int n,k,i,j,x,ans; 30 RD2(n,k); 31 REP(i,n) { 32 RD(x); 33 a[x]++; 34 } 35 x=0; 36 ans=0; 37 for(i=2000;i>=2;i--) { 38 if(a[i]>0) { 39 a[i]-=x; 40 int t=a[i]/k; 41 int tt=t*k; 42 int ttt=tt>=a[i]?t:t+1; 43 ans+=ttt*(i-1)*2; 44 x=ttt*k-a[i]; 45 //printf("%d %d %d\n",i,ttt,ans); 46 } 47 } 48 WN(ans); 49 return 0; 50 }
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。
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define ll __int64 14 #define usint unsigned int 15 #define mz(array) memset(array, 0, sizeof(array)) 16 #define minf(array) memset(array, 0x3f, sizeof(array)) 17 #define REP(i,n) for(int i=0;i<(n);i++) 18 #define FOR(i,x,n) for(int i=(x);i<=(n);i++) 19 #define RD(x) scanf("%d",&x) 20 #define RD2(x,y) scanf("%d%d",&x,&y) 21 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 22 #define WN(x) printf("%d\n",x); 23 #define RE freopen("D.in","r",stdin) 24 #define WE freopen("1.out","w",stdout) 25 26 char s[111111][2][55]; 27 28 int cmp(char a[55],char b[55]) { 29 int la=strlen(a),lb=strlen(b); 30 int l=min(la,lb); 31 int i; 32 for(i=0; i<l; i++) { 33 if(a[i]>b[i])return 1; 34 if(b[i]>a[i])return -1; 35 } 36 if(la>lb)return 1; 37 if(lb>la)return -1; 38 return 0; 39 } 40 41 int main() { 42 int n; 43 RD(n); 44 REP(i,n) { 45 scanf(" %s%s",s[i][0],s[i][1]); 46 //printf("%d %s!%s!\n",i,s[i][0],s[i][1]); 47 } 48 char q[55]; 49 int x,y=0,z,j; 50 RD(x); 51 y=x-1; 52 if(cmp(s[y][0],s[y][1])==-1) z=0; 53 else z=1; 54 bool flag=1; 55 REP(i,n-1) { 56 RD(x); 57 x--; 58 if(cmp(s[x][0],s[x][1])==-1) j=0; 59 else j=1; 60 if(cmp(s[x][j] , s[y][z])==-1){ 61 if(cmp(s[x][j^1], s[y][z])==-1){ 62 flag=0; 63 break; 64 }else{ 65 y=x; 66 z=j^1; 67 } 68 }else{ 69 y=x; 70 z=j; 71 } 72 } 73 if(flag)puts("YES"); 74 else puts("NO"); 75 return 0; 76 }
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
1 //#pragma comment(linker, "/STACK:102400000,102400000") 2 #include<cstdio> 3 #include<cmath> 4 #include<iostream> 5 #include<cstring> 6 #include<algorithm> 7 #include<cmath> 8 #include<map> 9 #include<set> 10 #include<stack> 11 #include<queue> 12 using namespace std; 13 #define ll __int64 14 #define usint unsigned int 15 #define mz(array) memset(array, 0, sizeof(array)) 16 #define mf1(array) memset(array, -1, sizeof(array)) 17 #define minf(array) memset(array, 0x3f, sizeof(array)) 18 #define REP(i,n) for(int i=0;i<(n);i++) 19 #define FOR(i,x,n) for(int i=(x);i<=(n);i++) 20 #define RD(x) scanf("%d",&x) 21 #define RD2(x,y) scanf("%d%d",&x,&y) 22 #define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z) 23 #define WN(x) printf("%d\n",x); 24 #define RE freopen("D.in","r",stdin) 25 #define WE freopen("1.out","w",stdout) 26 const int inf=2*1e9+7; 27 const int maxn=2222; 28 const int maxm=maxn*maxn; 29 30 int a[2222][2222]; 31 int n; 32 33 34 int bx[maxm],by[maxm],bz[maxm]; 35 int bn; 36 inline void add(const int &x,const int &y,const int &z){ 37 bx[bn]=x; 38 by[bn]=y; 39 bz[bn]=z; 40 bn++; 41 } 42 43 int d[maxm]; 44 bool cmp(const int &x,const int &y){ 45 return bz[x]<bz[y]; 46 } 47 48 int f[maxm]; 49 int gf(const int &x){ 50 if(f[x]==-1)return x; 51 f[x]=gf(f[x]); 52 return f[x]; 53 } 54 55 struct Edge{ 56 int next,v,l; 57 }e[maxm]; 58 int head[maxn],en; 59 60 inline void addedge(const int &x,const int &y,const int &z){ 61 e[en].v=y; 62 e[en].l=z; 63 e[en].next=head[x]; 64 head[x]=en; 65 en++; 66 } 67 68 int dis[maxn][maxn]; 69 70 inline void gank(const int &x,const int &y){ 71 int mi=inf; 72 for(int i=head[x];i!=-1;i=e[i].next){ 73 if(dis[y][e[i].v]!=-1){ 74 //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]); 75 mi=min(mi , e[i].l + dis[y][e[i].v]); 76 } 77 } 78 dis[y][x]=dis[x][y]=mi; 79 } 80 81 bool farm(){ 82 int i,j,k,x,y,z; 83 bn=0; 84 REP(i,n){ 85 REP(j,n){ 86 if(i!=j){ 87 if(a[i][j]==0)return 0; 88 }else{ 89 if(a[i][j]!=0)return 0; 90 } 91 } 92 } 93 REP(i,n){ 94 REP(j,i){ 95 if(a[i][j]!=a[j][i])return 0; 96 add(i,j,a[i][j]); 97 } 98 } 99 REP(i,bn){ 100 d[i]=i; 101 } 102 sort(d,d+bn,cmp); 103 mf1(f); 104 //mf1(head); 105 REP(i,n)REP(j,n)dis[i][j]=-1; 106 en=0; 107 mf1(head); 108 REP(i,n)dis[i][i]=0; 109 //en=0; 110 REP(i,bn){ 111 int q=d[i]; 112 int &x=bx[q],&y=by[q],&z=bz[q]; 113 int fx=gf(x),fy=gf(y); 114 if(fx!=fy){ 115 addedge(x,y,z); 116 addedge(y,x,z); 117 dis[x][y]=z; 118 dis[y][x]=z; 119 f[fx]=fy; 120 }else{ 121 if(dis[x][y]==-1) gank(x,y); 122 //printf("dis[%d][%d]=%d , z=%d\n",x,y,dis[x][y],z); 123 if(dis[x][y]!=z)return 0; 124 } 125 } 126 return 1; 127 } 128 129 130 131 int main() { 132 int i; 133 RD(n); 134 REP(i,n)REP(j,n) RD(a[i][j]); 135 if(farm())puts("YES"); 136 else puts("NO"); 137 return 0; 138 }
标签：des style blog http color io os ar for
原文地址：http://www.cnblogs.com/yuiffy/p/4004792.html
