[UOJ#404][CTSC2018]组合数问题(79分,提交答案题,模拟退火+匈牙利+DP)

1、4、5、6、10都是op=1的点，除4外直接通过模拟退火调参可以全部通过。

 #include<cmath>
 #include<ctime>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=;
 int n,m,K,op,ans,u,v,t[N][N],r[N][N],pos[N];
 struct E{ int u,v; }e[N];

 int sj(int l,int r){ return rand()%(r-l+)+l; }
 double Rand(){ return sj(,)/.; }

 int calc(){
     int res=;
     rep(i,,n) res+=t[i][pos[i]];
     rep(i,,m) res+=r[pos[e[i].u]][pos[e[i].v]];
     return res;
 }

 void SA(){
     for (double T=1e30; T>0.001; T*=0.99997){
         int p=sj(,n),q=sj(,K),ans1=ans-t[p][pos[p]]+t[p][q];
         rep(i,,m){
             if (e[i].u==p) ans1=ans1-r[pos[p]][pos[e[i].v]]+r[q][pos[e[i].v]];
             if (e[i].v==p) ans1=ans1-r[pos[e[i].u]][pos[p]]+r[pos[e[i].u]][q];
         }
         int delta=ans-ans1;
         if (delta> || Rand()<exp(delta/T)) ans=ans1,pos[p]=q;
     }
     rep(i,,){
         int p=sj(,n),q=sj(,K),ans1=ans-t[p][pos[p]]+t[p][q];
         rep(i,,m){
             if (e[i].u==p) ans1=ans1-r[pos[p]][pos[e[i].v]]+r[q][pos[e[i].v]];
             if (e[i].v==p) ans1=ans1-r[pos[e[i].u]][pos[p]]+r[pos[e[i].u]][q];
         }
         if (ans>ans1) ans=ans1,pos[p]=q;
     }
 }

 int main(){
     freopen("placement5.in","r",stdin);
     freopen("placement5.out","w",stdout);
     srand(time());
     scanf("%d%d%d%d",&n,&m,&K,&op);
     rep(i,,m) scanf("%d%d",&u,&v),e[i]=(E){u,v};
     rep(i,,n) rep(j,,K) scanf("%d",&t[i][j]);
     rep(i,,K) rep(j,,K) scanf("%d",&r[i][j]);
     rep(i,,n) pos[i]=; ans=calc(); SA();
     rep(i,,n) printf("%d ",pos[i]); puts("");
     return ;
 }

4号点是[1,133],[134,266],[267,399]三条链，做三次同样的DP即可。

 #include<cmath>
 #include<ctime>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=,inf=1e9;
 int n,m,K,op,ans,id,u,v,t[N][N],r[N][N],pre[N][N],f[N][N];

 void Print(int l,int i,int j){ if (i>l) Print(l,i-,pre[i][j]); printf("%d ",j); }

 int main(){
     freopen("placement4.in","r",stdin);
     freopen("placement4.out","w",stdout);
     srand(time());
     scanf("%d%d%d%d",&n,&m,&K,&op);
     rep(i,,m) scanf("%d%d",&u,&v);
     rep(i,,n) rep(j,,K) scanf("%d",&t[i][j]);
     rep(i,,K) rep(j,,K) scanf("%d",&r[i][j]);
     rep(i,,n) rep(j,,K) f[i][j]=inf;
     rep(i,,) rep(j,,K)
         rep(k,,K){
             int s=f[i-][k]+r[k][j]+t[i][j];
             if (s<f[i][j]) f[i][j]=s,pre[i][j]=k;
         }
     ans=inf;
     rep(i,,K) if (f[][i]<ans) ans=f[][i],id=i;
     Print(,,id);
     rep(i,,K) f[][i]=t[][i];
     rep(i,,) rep(j,,K)
         rep(k,,K){
             int s=f[i-][k]+r[k][j]+t[i][j];
             if (s<f[i][j]) f[i][j]=s,pre[i][j]=k;
         }
     ans=inf;
     rep(i,,K) if (f[][i]<ans) ans=f[][i],id=i;
     Print(,,id);
     rep(i,,K) f[][i]=t[][i];
     rep(i,,) rep(j,,K)
         rep(k,,K){
             int s=f[i-][k]+r[k][j]+t[i][j];
             if (s<f[i][j]) f[i][j]=s,pre[i][j]=k;
         }
     ans=inf;
     rep(i,,K) if (f[][i]<ans) ans=f[][i],id=i;
     Print(,,id);
     return ;
 }

2、3、8、9同样可以用模拟退火做，发现题目给的simulator会创建一个rex.txt来存放op=2时的答案，于是直接用它做估价函数即可。这样第二个点可以得到满分，第3个点可以得到3分，第8个点可以得到1分，第9个点可以得到5分。

 #include<cmath>
 #include<ctime>
 #include<cstdio>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=;
 int n,m,K,op,ans,u,v,t[N][N],r[N][N],pos[N];
 struct E{ int u,v; }e[N*N];

 int sj(int l,int r){ return rand()%(r-l+)+l; }
 double Rand(){ return sj(,)/.; }

 int calc(){
     freopen("placement9.out","w",stdout);
     rep(i,,n) printf("%d ",pos[i]); puts("");
     fclose(stdout);
     system("./simulator placement9.in placement9.out");
     freopen("res.txt","r",stdin);
     int res; scanf("%d",&res); fclose(stdin); return res;
 }

 void SA(){
     for (double T=1e10; T>0.001; T*=0.997,cerr<<T<<endl){
         int p=sj(,n),q=sj(,K),w=pos[p]; pos[p]=q;
         int ans1=calc(),delta=ans-ans1;
         if (delta> || Rand()<exp(delta/T)) ans=ans1; else pos[p]=w;
     }
     rep(i,,){
         int p=sj(,n),q=sj(,K),w=pos[p]; pos[p]=q;
         int ans1=calc();
         if (ans>ans1) ans=ans1; else pos[p]=w;
     }
 }

 int main(){
     freopen("placement9.in","r",stdin);
     srand(time());
     scanf("%d%d%d%d",&n,&m,&K,&op);
     rep(i,,m) scanf("%d%d",&u,&v),e[i]=(E){u,v};
     rep(i,,n) rep(j,,K) scanf("%d",&t[i][j]);
     rep(i,,K) rep(j,,K) scanf("%d",&r[i][j]);
     rep(i,,n) pos[i]=; ans=calc(); SA();
     freopen("placement9.out","w",stdout);
     rep(i,,n) printf("%d ",pos[i]); puts("");
     return ;
 }

7号点可以直接跑匈牙利得到结果。

 #include<cmath>
 #include<ctime>
 #include<cstdio>
 #include<cstdlib>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=;
 int n,m,K,op,u,v,vis[N],lnk[N],ans[N],t[N][N],r[N][N],pos[N];

 bool work(int x,int p){
     rep(i,,K) if (t[x][i]<= && vis[i]!=p){
         vis[i]=p;
         if (lnk[i]==- || work(lnk[i],p)){ lnk[i]=x; return ; }
     }
     return ;
 }

 int main(){
     freopen("placement7.in","r",stdin);
     freopen("placement7.out","w",stdout);
     srand(time());
     scanf("%d%d%d%d",&n,&m,&K,&op);
     rep(i,,m) scanf("%d%d",&u,&v);
     rep(i,,n) rep(j,,K) scanf("%d",&t[i][j]);
     rep(i,,K) rep(j,,K) scanf("%d",&r[i][j]);
     rep(i,,K) lnk[i]=-;
     rep(i,,n) work(i,i);
     rep(i,,K) if (~lnk[i]) ans[lnk[i]]=i;
     rep(i,,n) printf("%d ",ans[i]);
     return ;
 }