找最长的其实是很裸的状态压缩DP，棘手的地方是要统计数量，其实只要再来一个数组存就好。
不过代码比较长，细节要注意的地方毕较多，wa了很多发，还是要仔细啊
用递推和记忆化搜索分别写了一遍
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
using namespace std;
typedef long long LL;
const int maxn = 14;
const int INF = INT_MAX / 3;
int n,m,dist[maxn][maxn];
LL v[maxn];
LL f[maxn][maxn][1 << 14];
LL cnt[maxn][maxn][1 << 14];
void solve() {
memset(f,-1,sizeof(f));
memset(cnt,0,sizeof(cnt));
int maxstate = (1 << n) - 1; //init
for(int i = 0;i < n;i++) {
for(int j = 0;j < n;j++) if(dist[i][j]) {
int s = (1 << i) | (1 << j);
f[i][j][s] = v[i] * v[j] + v[i] + v[j];
cnt[i][j][s] = 1;
}
}
//dp
LL ans = 0,ct = 0;
for(int s = 0;s <= maxstate;s++) {
for(int i = 0;i < n;i++) if(s & (1 << i)) {
for(int j = 0;j < n;j++) if(dist[i][j] && (s & (1 << j)) && f[i][j][s] != -1) {
for(int k = 0;k < n;k++) if(dist[j][k] && !(s & (1 << k))) {
int ns = s | (1 << k);
LL &nowstate = f[i][j][s],&nextstate = f[j][k][ns];
LL &nowcnt = cnt[i][j][s],&nextcnt = cnt[j][k][ns];
LL addv = v[j] * v[k] + v[k];
if(dist[i][k]) addv += v[i] * v[j] * v[k];
if(nowstate + addv > nextstate) {
nextstate = nowstate + addv;
nextcnt = nowcnt;
}
else if(nowstate + addv == nextstate) {
nextcnt += nowcnt;
}
}
if(s == maxstate) {
if(f[i][j][s] > ans) {
ans = f[i][j][s];
ct = cnt[i][j][s];
}
else if(f[i][j][s] == ans) {
ct += cnt[i][j][s];
}
}
}
}
}
if(ans == 0) cout << "0 0" << endl;
else cout << ans << " " << ct / 2 << endl;
}
int main() {
//freopen("in.txt","r",stdin);
int T; scanf("%d",&T);
while(T--) {
memset(dist,0,sizeof(dist));
scanf("%d%d",&n,&m);
for(int i = 0;i < n;i++) cin >> v[i];
for(int i = 0;i < m;i++) {
int a,b; cin >> a >> b;
a--; b--;
dist[a][b] = dist[b][a] = 1;
}
if(n == 1) cout << v[0] << " " << 1 << endl;
else solve();
}
return 0;
}
1 #include <cstdio>
2 #include <sstream>
3 #include <fstream>
4 #include <cstring>
5 #include <iostream>
6 #include <algorithm>
7 #include <map>
8 #include <cctype>
9 #include <ctime>
10 #include <set>
11 #include <climits>
12 #include <vector>
13 #include <queue>
14 #include <stack>
15 #include <cstdlib>
16 #include <cmath>
17 #include <string>
18 #include <list>
19
20 #define INPUT_FILE "in.txt"
21 #define OUTPUT_FILE "out.txt"
22
23 using namespace std;
24
25 typedef long long LL;
26 const int INF = INT_MAX / 2;
27
28 void setfile() {
29 freopen(INPUT_FILE,"r",stdin);
30 freopen(OUTPUT_FILE,"w",stdout);
31 }
32
33 const int maxn = 13;
34
35 int n,maxstate;
36 LL w[maxn][maxn],c[maxn];
37 LL note[maxn][maxn][1 << 13];
38 LL cnt[maxn][maxn][1 << 13];
39
40 LL dfs(int prev,int now,int state) {
41 if(state == maxstate) return note[prev][now][state] = 0;
42 LL ret = -1,nret;
43 if(note[prev][now][state] != -1) {
44 return note[prev][now][state];
45 }
46 for(int i = 0;i < n;i++) {
47 if((state & (1 << i)) == 0 && w[now][i] != -1) {
48 LL addv = w[now][i] + c[i];
49 if(w[prev][i] != -1) addv += c[i] * c[prev] * c[now];
50 nret = dfs(now,i,state | (1 << i)) + addv;
51 if(nret > ret) {
52 ret = nret;
53 }
54 }
55 }
56 if(ret == -1) return -INF * 100000LL;
57 return note[prev][now][state] = ret;
58 }
59
60 LL getcount(int prev,int now,int state) {
61 if(state == maxstate) return 1;
62 if(cnt[prev][now][state] != -1) return cnt[prev][now][state];
63 LL ret = 0;
64 for(int i = 0;i < n;i++) {
65 if((state & (1 << i)) == 0 && w[now][i] != -1) {
66 LL addv = w[now][i] + c[i];
67 if(w[prev][i] != -1) addv += c[i] * c[prev] * c[now];
68 if(note[prev][now][state] == note[now][i][state | (1 << i)] + addv) {
69 ret += getcount(now,i,state | (1 << i));
70 }
71 }
72 }
73 return cnt[prev][now][state] = ret;
74 }
75
76 int main() {
77 int T;
78 cin >> T;
79 while(T--) {
80 memset(w,-1,sizeof(w));
81 memset(cnt,-1,sizeof(cnt));
82 memset(note,-1,sizeof(note));
83 int m;
84 cin >> n >> m;
85 for(int i = 0;i < n;i++) {
86 cin >> c[i];
87 }
88 for(int i = 0;i < m;i++) {
89 int u,v;
90 cin >> u >> v;
91 u--; v--;
92 w[u][v] = w[v][u] = c[u] * c[v];
93 }
94 maxstate = (1 << n) - 1;
95 LL ans = 0,ccnt = 0;
96 for(int i = 0;i < n;i++) {
97 for(int j = 0;j < n;j++) if(i != j && w[i][j] != -1) {
98 LL ret = dfs(i,j,(1<<i)|(1<<j)) + w[i][j] + c[i] + c[j];
99 ans = max(ans,ret);
100 }
101 }
102 for(int i = 0;i < n;i++) {
103 for(int j = 0;j < n;j++) {
104 if(note[i][j][(1 << i)|(1 << j)] + w[i][j] + c[i] + c[j] == ans) {
105 ccnt += getcount(i,j,(1<<j)|(1<<i));
106 }
107 }
108 }
109 if(n == 1) cout << c[0] << " " << 1 << endl;
110 else if(ccnt == 0) cout << "0 0" << endl;
111 else cout << ans << " " << ccnt / 2 << endl;
112 }
113 return 0;
114 }
POJ 2288 Islands and Bridges 哈密尔顿路 状态压缩DP
时间： 2024-12-28 23:30:14