• 【bzoj4806~bzoj4809】 象棋四连发 DP-高精度-匈牙利算法-dfs


    都是经典题了吧。。我好无聊。。

    4806

    4806-1801是双倍经验。。DP方程看代码吧。。

     1 /* http://www.cnblogs.com/karl07/ */
     2 #include <cstdlib>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <cmath>
     6 #include <algorithm>
     7 using namespace std;
     8 
     9 #define ll long long 
    10 //#define P 9999973 //bzoj1801
    11 #define P 999983 //bzoj4806
    12 #define C(x) ((x)*(x-1)/2)
    13 int n,m;
    14 ll f[105][105][105];
    15 
    16 int main(){
    17     scanf("%d%d",&n,&m);
    18     f[0][0][0]=1;
    19     for (int i=1;i<=n;i++){
    20         for (int j=0;j<=m;j++){
    21             for (int k=0;k<=m-j;k++){
    22                 f[i][j][k]+=f[i-1][j][k];
    23                 if (j>=1)          f[i][j][k]+=f[i-1][j-1][k]*(m-j-k+1);
    24                 if (k>=1 && j<=m-1)f[i][j][k]+=f[i-1][j+1][k-1]*(j+1);
    25                 if (j>=2)          f[i][j][k]+=f[i-1][j-2][k]*C(m-j-k+2);
    26                 if (k>=2 && j<=m-2)f[i][j][k]+=f[i-1][j+2][k-2]*C(j+2);
    27                 if (j>=1 && k>=1)  f[i][j][k]+=f[i-1][j][k-1]*j*(m-j-k+1);
    28                 f[i][j][k]%=P;
    29             }
    30         }
    31     }
    32     ll ans=0;
    33     for (int i=0;i<=m;i++){
    34         for (int j=0;j<=m-i;j++){
    35             ans+=f[n][i][j];
    36             ans%=P;
    37         }
    38     }
    39     printf("%lld
    ",ans);
    40     return 0;
    41 }

    4807

    就是 max(m,n) C min (m,n) 

     1 /* http://www.cnblogs.com/karl07/ */
     2 #include <cstdlib>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <cmath>
     6 #include <algorithm>
     7 using namespace std;
     8 struct INT {
     9     int b,a[60];
    10     INT (){b=0; for (int i=0;i<=50;i++) a[i]=0;}
    11 };
    12 int A,B,c,n,m;
    13 int P[1000005],p[1000005],cnt[1000005],ip[1000005];
    14 INT Int(int x){
    15     INT a;
    16     for (a.b=0 ; x ; a.a[++a.b]=x%10,x/=10);
    17     return a;
    18 }
    19 
    20 INT operator * (const INT &x,const INT &y){
    21     INT a;
    22     a.b=min(50,x.b+y.b-1);
    23     for (int i=1;i<=x.b;i++) for (int j=1;j<=y.b;j++) if (i+j-1<=50) a.a[i+j-1]+=x.a[i]*y.a[j]; 
    24     for (int i=1;i<=a.b;a.b+=(i==a.b && a.a[i+1] && i+1<=50),i++)  a.a[i+1]+=a.a[i]/10,a.a[i]%=10;
    25     for (;a.a[a.b]==0 && a.b!=1; a.b--);
    26     return a;
    27 }
    28 
    29 void print_INT(INT x){
    30     for (int i=min(x.b,50);i>=1;i--) printf("%d",x.a[i]); 
    31     puts("");
    32 }
    33 
    34 void Prime(int n){
    35     for (int i=2;i<=n;i++){
    36         if (!P[i]){
    37             p[++c]=i,ip[i]=c;
    38             for (int j=i+i;j<=n;j+=i) P[j]=1;
    39         }
    40     }
    41 }
    42 
    43 void pt(int x) {
    44     printf("%d
    ",x);
    45 }
    46 
    47 void fac(int x,int y){
    48     for (int i=1;i<=c && P[x];i++) while (!(x%p[i])) x/=p[i],cnt[i]+=y;
    49     cnt[ip[x]]+=y;
    50 }
    51 
    52 INT ans=Int(1);
    53 int main(){
    54     scanf("%d%d",&A,&B);
    55     n=max(A,B),m=max(max(A,B)-min(A,B),min(A,B));
    56     Prime(n);
    57     for (int i=m+1;i<=n;i++) fac(i,1);
    58     for (int i=2;i<=n-m;i++) fac(i,-1);
    59     for (int i=1;i<=c;i++) while (cnt[i]--) ans=ans*Int(p[i]);
    60     print_INT(ans);
    61     return 0;
    62 }

    4808

    黑白染色然后匈牙利

     1 /* http://www.cnblogs.com/karl07/ */
     2 #include <cstdlib>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <cmath>
     6 #include <algorithm>
     7 using namespace std;
     8 
     9 struct edge{
    10     int next,to;
    11 }e[1000005];
    12 int ade,n,m,cnt,T,c;
    13 int x[]={1,2,-1,-2,-1,2,1,-2};
    14 int y[]={2,1,-2,-1,2,-1,-2,1};
    15 int mp[205][205],id[205][205],first[1000005],vis[1000005],match[1000005];
    16 
    17 void addedge(int x,int y){
    18     e[++ade].next=first[x];
    19     e[ade].to=y;
    20     first[x]=ade;
    21 }
    22 
    23 bool check(int x,int y){
    24     return (x>=1 && x<=n && y>=1 && y<=m && !mp[x][y]);
    25 }
    26 
    27 #define s e[x].to
    28 bool hungary(int p){
    29     for (int x=first[p];x;x=e[x].next){
    30         if (vis[s]!=T){
    31             vis[s]=T;
    32             if (!match[s] || hungary(match[s])) {match[s]=p; return 1;}
    33         }
    34     }
    35     return 0;
    36 }
    37 #undef s
    38 
    39 int main(){
    40     scanf("%d%d",&n,&m);
    41     for (int i=1;i<=n;i++){
    42         for (int j=1;j<=m;j++){
    43             scanf("%d",&mp[i][j]);
    44             cnt+=(!mp[i][j]);
    45             id[i][j]=++c;
    46         }
    47     }
    48     for (int i=1;i<=n;i++)
    49         for (int j=1;j<=m;j++)
    50             if (((i+j)&1) && !mp[i][j])for (int k=0;k<8;k++) if (check(i+x[k],j+y[k])) addedge(id[i][j],id[i+x[k]][j+y[k]]);
    51     for (int i=1;i<=n;i++){
    52         for (int j=1;j<=m;j++){
    53             if (!mp[i][j] && ((i+j)&1)){
    54                 T=id[i][j];
    55                 cnt-=hungary(T);
    56             }
    57         }
    58     }
    59     printf("%d
    ",cnt);
    60     return 0;
    61 }

    4809

    直接dfs

     1 /* http://www.cnblogs.com/karl07/ */
     2 #include <cstdlib>
     3 #include <cstdio>
     4 #include <cstring>
     5 #include <cmath>
     6 #include <algorithm>
     7 using namespace std;
     8 
     9 int n,ans;
    10 int ban[20][20],y[20],a[40],b[40];
    11 
    12 void dfs(int i){
    13     if (i>n) {ans++; return;} 
    14     for(int j=1;j<=n;j++) if (!(ban[i][j] || y[j] || a[i+j-1] || b[n-i+j] )){
    15         y[j]=a[i+j-1]=b[n-i+j]=1;
    16         dfs(i+1);
    17         y[j]=a[i+j-1]=b[n-i+j]=0;        
    18     }
    19 }
    20 
    21 int main(){
    22     scanf("%d",&n);
    23     for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) scanf("%d",&ban[i][j]);
    24     dfs(1);
    25     printf("%d
    ",ans);
    26     return 0;
    27 }
    皇后
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  • 原文地址:https://www.cnblogs.com/karl07/p/6746312.html
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