hdu3001(状态压缩dp）

题目链接：点击打开链接

题目：

   After coding so many days,Mr Acmer wants to have a good rest.So travelling is the best choice!He has decided to visit n cities(he insists on seeing all the cities!And he does not mind which city being his start station because superman can bring him to any
city at first but only once.), and of course there are m roads here,following a fee as usual.But Mr Acmer gets bored so easily that he doesn't want to visit a city more than twice!And he is so mean that he wants to minimize the total fee!He is lazy you see.So
he turns to you for help.
分析：
   每天路径最多访问两次，可以使用三进制状态压缩，d[S][j] 表示在状态s下最后一个访问j时的最小花费;
   状态转移方程：d[S][k] = min(d[S][k], d[S-state[k]][j] + G[j][k]);    不过这里通过状态d[S][j] 转移到 ---> d[S+state[k]][k] = min(d[S+state[k]][k], d[S][j]+G[j][k]);

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int maxn = 11;
int n, m;
int G[60000][maxn], state[11], vis[60000][maxn];
const int INF = 0x3f3f3f3f;
int d[60000][maxn];

void init(){
state[0] = 1;
for(int i = 1; i <= 10; ++i){
state[i] = 3*state[i-1];
}
for(int i = 0; i < state[10]; ++i){
int tmp = i;
for(int j = 0; j < 10; ++j){
vis[i][j] = tmp%3;
tmp /= 3;
}
}
}
int deal(){
memset(d, 0x3f, sizeof(d));
for(int i = 0; i < n; ++i) d[state[i]][i] = 0;
for(int i = 1; i < state
; ++i){
for(int j = 0; j < n; ++j){
if(d[i][j] == INF) continue;
for(int k = 0; k < n; ++k){
if(k==j) continue;
if(vis[i][k] < 2 && G[j][k]!=INF){
d[i+state[k]][k] = min(d[i+state[k]][k], d[i][j]+G[j][k]);
}
}
}
}
int ans = INF;
for(int i = 1; i < state
; ++i){
int ok = 1;
for(int j = 0; j < n; ++j){
if(vis[i][j] == 0){
ok = 0; break;
}
}
if(ok){
for(int j = 0; j < n; ++j){
ans = min(ans, d[i][j]);
}
}
}
return ans;
}

int main()
{
init();
while(~scanf("%d%d", &n, &m)){
int u, v, w;
memset(G, 0x3f, sizeof(G));
for(int i = 0; i < m; ++i){
scanf("%d%d%d", &u, &v, &w);
u--, v--;
if(w < G[u][v]){
G[u][v] = G[v][u] = w;
}
}
int ans = deal();
if(ans == INF) ans = -1;
printf("%d\n", ans);
}
}