Uva 116 - Unidirectional TSP

 3Y today

( 10Y before )

终于搞定了!!!

这玩意好烦

估计是之前字典序没弄好==

dp的构建就是分段图遍历的做法

然后解释一下ok

用于字典序输出

反着找出一切可能被遍历的点

然后正向输出

没输出一个值

把所有不可能的点排除( ok清零 )

就好了

终于搞定了!!!!!

#include<stdio.h>#define LL long longLL map[20][110];LL dp[20][110];LL m,n;LL min3(LL a,LL b,LL c){LL res=a;if(b<res)res=b;if(c<res)res=c;return res;}int main(){while(scanf("%lld%lld",&n,&m)!=EOF){//n行m列 int i,j;for(i=1;i<=n;i++)for(j=1;j<=m;j++)scanf("%lld",&map[i][j]);for(i=1;i<=m;i++){for(j=1;j<=n;j++){dp[j][i]=map[j][i];if(i!=1){int x,y;x=j+1;if(x==n+1)x=1;y=j-1;if(y==0)y=n;dp[j][i]+=min3(dp[x][i-1],dp[j][i-1],dp[y][i-1]);}}}/*for(i=1;i<=n;i++){for(j=1;j<=m;j++)printf("%d ",dp[i][j]);printf("\n");}*/LL res=dp[n][m],pr=n;for(i=1;i<=n;i++)if(res>dp[i][m])res=dp[i][m],pr=i;bool ok[20][110];for(i=1;i<=n;i++)for(j=1;j<=m;j++)ok[i][j]=0;for(j=m;j>=1;j--){for(i=1;i<=n;i++){if(j==m){if(dp[i][j]==res)ok[i][j]=1;}else {int k;for(k=1;k<=n;k++){if(ok[k][j+1]){LL aim=dp[k][j+1]-map[k][j+1];int x,y;x=k+1;if(x==n+1)x=1;y=k-1;if(y==0)y=n;if(dp[x][j]==aim)ok[x][j]=1;if(dp[k][j]==aim)ok[k][j]=1;if(dp[y][j]==aim)ok[y][j]=1;}}}}}for(j=1;j<=m;j++){for(i=1;i<=n;i++){if(ok[i][j]){if(j!=1)printf(" ");printf("%d",i);int k;int x,y;x=i+1;if(x==n+1)x=1;y=i-1;if(y==0)y=n;for(k=1;k<=n;k++){if(k==i||k==x||k==y){if(dp[k][j+1]-map[k][j+1]!=dp[i][j])ok[k][j+1]=0;}else ok[k][j+1]=0;}break;}}}printf("\n%lld\n",res);}return 0;}