bzoj 3055礼物运送 floyed + 状压DP

bzoj 3055: 礼物运送

floyed first 设f[i][S]表示取到了S集合中的所有点（不一定是经过的所有点），最后停在了i的最优值。

初始就f[i][{i}] = dis[1][i]

状态转移直接转就好了

f[i][S] + dis[i][j] -> f[j][S + {j}]其中 i 属于 S,1 <= j <= n

设tim[S] = min{f[i][S]}

答案就取到

           ans = min{max{tim[S],tim[Cs]}}

#include <map>
#include <queue>
#include <cstdio>
#include <cstring>
#include <climits>
#include <algorithm>
using namespace std;
inline void read(int &x){
    x=0;char ch;bool flag = false;
    while(ch=getchar(),ch<'!');if(ch == '-') ch=getchar(),flag = true;
    while(x=10*x+ch-'0',ch=getchar(),ch>'!');if(flag) x=-x;
}
inline int cat_min(const int &a,const int &b){return a<b ? a:b;}
inline int cat_max(const int &a,const int &b){return a>b ? a:b;}
inline int cat_abs(const int &x){return x < 0 ? -x : x;}
const int maxn = 19;
const int maxs = (1<<19) + 1;
int dis[maxn][maxn],n,f[maxn][maxs],tim[maxs];
inline void init(){
    memset(dis,0x3f,sizeof dis);
    memset(f,0x3f,sizeof f);
    memset(tim,0x3f,sizeof tim);
}
void floyed(){
    for(int i=1;i<=n;++i){
        dis[i][i] = 0;
        for(int j=1;j<=n;++j){
            if(i == j) continue;
            for(int k=1;k<=n;++k){
                dis[i][j] = cat_min(dis[i][j],
                dis[i][k] + dis[k][j]);
            }
        }
    }return;
}
map<int,int>ma;
int main(){
 
    int m;read(n);read(m);
    init();
    for(int i=1,u,v,d;i<=m;++i){
        read(u);read(v);read(d);
        if(dis[u][v] > d) dis[u][v] = dis[v][u] = d;
    }
    floyed();
    for(int i=0;i<=18;++i) ma[1<<i] = i+1;
    for(int i=1;i<=n;++i) f[i][(1<<0) | (1<<(i-1))] = dis[1][i];
    for(int i=0,x=0;i<(1<<n);x = (++i) ){
        while(x){
            int p = ma[x&-x];x -=x&-x;
            for(int j=1;j<=n;++j){
                if( f[j][i | (1<<(j-1))] > f[p][i] + dis[p][j]){
                    f[j][i | (1<<(j-1))] = f[p][i] + dis[p][j];
                }
            }
        }
    }
    for(int i=0,x=0;i<(1<<n);x = (++i) ){
        while(x){
            int p = ma[x & -x];x -= x&-x;
            tim[i]=cat_min(tim[i],f[p][i]);
        }
    }
    int ans = 0x7f7f7f7f;
    for(int i=0;i<(1<<n);++i){
        ans = cat_min(ans,cat_max(tim[i],
        tim[1 | (( (1 << n)-1 )^i) ]));
    }printf("%d
",ans);
    //getchar();getchar();
    fclose(stdin);fclose(stdout);
    return 0;
}