HDU 3681 Prison Break 哈密顿回路DP(二)

2014-11-24 07:44:33 · 作者: · 浏览: 1
",ans);
}
return 0;
}
DFS做法(用时31ms)
[cpp]
#include
#include
#include
using namespace std;
struct Edge{
int v,next;
}edge[2000];
int head[300],cnt,ids,fin,mid;
int n,m;
char map[20][20];
int id[20][20];
int c[20][2];
int dis[20][20];
struct Node{
int s,d;
};
void init(){
memset(head,-1,sizeof(head));
memset(id,-1,sizeof(id));
memset(dis,-1,sizeof(dis));
cnt=0,ids=1,fin=0;
}
void addedge(int u,int v){
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
}
void bfs(int s){
int i,j,u,v;
struct Node tem,tem1;
queueq;
bool ok[400];
memset(ok,0,sizeof(ok));
tem.s=s,tem.d=0;
q.push(tem);
ok[s]=1;
while(!q.empty()){
tem=q.front();
q.pop();
u=tem.s;
if(id[u/m][u%m]!=-1) dis[id[s/m][s%m]][id[u/m][u%m]]=tem.d;
for(i=head[u];i!=-1;i=edge[i].next){
v=edge[i].v;
if(ok[v])continue;
ok[v]=1;
tem1.s=v;
tem1.d=tem.d+1;
q.push(tem1);
}
}
}
int has[20];
int dfs(int pos,int len,int sta){
if((sta&fin)==fin)return 1;
for(int i=0;i
if(has[i] || dis[pos][i]==-1)continue;
if(len>=dis[pos][i]){
if(map[c[i][0]][c[i][1]]=='G'){
has[i]=1;
if(dfs(i,mid,sta|(1<
has[i]=0;
}
else{
has[i]=1;
if(dfs(i,len-dis[pos][i],sta|(1<
has[i]=0;
}
}
}
return 0;
}
int main(){
int i,j,k;
while(scanf("%d %d",&n,&m) && n!=0 && m!=0){
init();
for(i=0;i
scanf("%s",map[i]);
for(i=0;i
for(j=0;j
if(map[i][j]=='D')continue;
if(map[i][j]=='F'){
c[0][0]=i;
c[0][1]=j;
id[i][j]=0;
}
else if(map[i][j]=='G'){
c[ids][0]=i;
c[ids][1]=j;
id[i][j]=ids++;
}
else if(map[i][j]=='Y'){
c[ids][0]=i;
c[ids][1]=j;
id[i][j]=ids++;
fin|=(1<<(ids-1));
}
if(i-1>=0 && map[i-1][j]!='D')
addedge(i*m+j,(i-1)*m+j);
if(i+1
addedge(i*m+j,(i+1)*m+j);
if(j-1>=0 && map[i][j-1]!='D')
addedge(i*m+j,i*m+j-1);
if(j+1
addedge(i*m+j,i*m+j+1);
}
for(i=0;i
bfs(c[i][0]*m+c[i][1]);
int flag=1;
for(i=1;i
if(map[c[i][0]][c[i][1]]=='Y')
if(dis[0][i]==-1){
flag=0;
break;
}
}
int ans=-1;
if(flag){
int l=0,r=n*m*(ids-1);
while(l<=r){
mid=(l+r)>>1;
memset(has,0,sizeof(has));
has[0]=1; www.2cto.com
if(dfs(0,mid,1)){
ans=mid;
r=mid-1;
}
else
l=mid+1;
}
}
printf("%d\n",ans);
}
return