P4802 [CCO 2015]路短最

${\color{Cyan}{>>Question}}$

简单的状压$dp$,

和P1171 售货员的难题简直一模一样(但也正因为此把我坑了)

令$f[i][j]$表示状态$i$,最后在$j$的最短路最长路(又坑了我)

有$$f[i][j] = max\left\{f[i-\left \{ j \right \}][k]+dis[k][j]\;\;|\;\;k\in (i-\left \{ j \right \})\right\}$$

!!要初始化为极小值!!(不初始化只有$46pts$)

!!边界!!$$f[i][0] = 0$$

枚举时可以只枚举奇数的状态(因为$0$号点必选)

上代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std; 

template <typename T> void in(T &x) {
    x = 0; T f = 1; char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') f = -1; ch = getchar();}
    while( isdigit(ch)) {x = 10 * x + ch - 48; ch = getchar();}
    x *= f;
}

template <typename T> void out(T x) {
    if(x < 0) x = -x , putchar('-');
    if(x > 9) out(x/10);
    putchar(x%10 + 48);
}
//-------------------------------------------------------

const int N = 20;

int n,m;
int a[N][N];
int f[1<<N][N];

int main() {
    int i,j,k,u,v;
    in(n); in(m);
    for(i = 0;i < m; ++i) {
        in(u); in(v); in(a[u][v]);
    }
    memset(f,~0x3f,sizeof(f));//debug 要赋极小值
    f[1][0] = 0;//!边界! 
    for(i = 3;i < (1<<n); i += 2) {
        f[i][0] = 0;//!边界! 
        for(j = 0;j < n; ++j) {
            if(!(i&(1<<j))) continue;
            for(k = 0;k < n; ++k) {
                if(!a[k][j] || !(i&(1<<k))) continue;
                f[i][j] = max(f[i][j],f[i^(1<<j)][k]+a[k][j]);
            }
        }
    }
    //out(f[(1<<n)-1][n-1]);//debug
    int ans = 0;
    for(i = 1|(1<<n-1);i < (1<<n); i += 2) ans = max(ans,f[i][n-1]);
    out(ans);
    return 0;
}