很经典的一道状压dp(似乎叫做旅行商问题)，用f[i][s]表示在到达点i，已经经过的城市用二进制表示为s，于是方程就很简单了:
f[i][s] = min { f[j][s ^ (1 << j)] + dis[j][i]| s & (1 << j) != 0}
然后用记忆化搜索即可，注意方向，因为dis[i][j]可能不等于dis[j][i]。（下面的代码某个处理似乎没有必要）
Code
1 #include<iostream>
2 #include<cstdio>
3 #include<cctype>
4 #include<cstring>
5 #include<cstdlib>
6 #include<fstream>
7 #include<sstream>
8 #include<algorithm>
9 #include<map>
10 #include<set>
11 #include<queue>
12 #include<vector>
13 #include<stack>
14 using namespace std;
15 typedef bool boolean;
16 #define INF 0xfffffff
17 #define smin(a, b) a = min(a, b)
18 #define smax(a, b) a = max(a, b)
19 template<typename T>
20 inline void readInteger(T& u){
21 char x;
22 int aFlag = 1;
23 while(!isdigit((x = getchar())) && x != ‘-‘);
24 if(x == ‘-‘){
25 x = getchar();
26 aFlag = -1;
27 }
28 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
29 ungetc(x, stdin);
30 u *= aFlag;
31 }
32
33 int n;
34 int dis[16][16];
35 int f[16][(1 << 17)];
36 boolean vis[16][(1 << 17)];
37
38 inline void init() {
39 readInteger(n);
40 for(int i = 1; i <= n; i++)
41 for(int j = 1; j <= n; j++)
42 readInteger(dis[i][j]);
43 for(int i = 1; i <= n; i++)
44 dis[i][0] = dis[i][1], dis[0][i] = dis[1][i];
45 }
46
47 int dfs(int local, int status) {
48 if(vis[local][status]) return f[local][status];
49 vis[local][status] = true;
50 for(int i = 0; i <= n; i++) {
51 if(status & (1 << i)) {
52 int ret = dfs(i, status ^ (1 << i));
53 smin(f[local][status], f[i][status ^ (1 << i)] + dis[i][local]);
54 }
55 }
56 return f[local][status];
57 }
58
59 inline void solve() {
60 memset(vis, false, sizeof(vis));
61 memset(f, 0x7f, sizeof(f));
62 vis[1][0] = true;
63 f[1][0] = 0;
64 int res = dfs(0, (1 << (n + 1)) - 2);
65 printf("%d", res);
66 }
67
68 int main() {
69 freopen("salesman.in", "r", stdin);
70 freopen("salesman.out", "w", stdout);
71 init();
72 solve();
73 return 0;
74 }
因为关灯不耗时间，所以从一个地方走到另一个地方，从贪心的角度来讲，肯定要把沿路的灯都关掉，因此得到原问题转化成了区间dp。当然还要考虑是从做走到右还是从右走到左，当然可以直接用f[i][j]表示，但是为了防止各种手抽手贱导致半天调不出来，还是加了一维[0/1]，表示在从右走到左(0)还是从左走到右。于是方程很容易就出来了，详细的看代码，这里就简单地写了，从f[i][j]转移到f[i - 1][j]或者f[i][j + 1]，然后加上路程乘以未被关掉的所有灯的功率。
至于这个功率可以用前缀和先预处理出，接着用总功率减去这一段的功率就行了。
由于开始以为灯的位置不是有序的，特意排了道序，可以无视。
Code
#include<iostream>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<cstdlib>
#include<fstream>
#include<sstream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
using namespace std;
typedef bool boolean;
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif
#define INF 0xfffffff
#define smin(a, b) a = min(a, b)
#define smax(a, b) a = max(a, b)
template<typename T>
inline void readInteger(T& u){
char x;
int aFlag = 1;
while(!isdigit((x = getchar())) && x != ‘-‘);
if(x == ‘-‘){
x = getchar();
aFlag = -1;
}
for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
ungetc(x, stdin);
u *= aFlag;
}
typedef class tower {
public:
int pos;
int w;
int index;
tower(const int pos = 0, const int w = 0, const int index = 0):pos(pos), w(w), index(index) { }
boolean operator < (tower a) const {
return pos < a.pos;
}
}tower;
int n, c, rc;
long long f[2][1005][1005];
tower tows[1005];
long long sumw[1005];
inline void init() {
readInteger(n);
readInteger(c);
for(int i = 1; i <= n; i++) {
readInteger(tows[i].pos);
readInteger(tows[i].w);
tows[i].index = i;
}
sort(tows + 1, tows + n + 1);
sumw[0] = 0;
for(int i = 1; i <= n; i++) {
sumw[i] = sumw[i - 1] + tows[i].w;
if(tows[i].index == c) rc = i;
}
}
inline void solve() {
memset(f, 0x37, sizeof(f));
f[0][rc][rc] = f[1][rc][rc] = 0;
if(rc > 1)
f[0][rc - 1][rc] = (tows[rc].pos - tows[rc - 1].pos) * (sumw[n] - tows[rc].w);
if(rc < n)
f[1][rc][rc + 1] = (tows[rc + 1].pos - tows[rc].pos) * (sumw[n] - tows[rc].w);
for(int l = 1; l < n; l++) {
for(int i = 1; i + l <= n; i++) {
int j = i + l;
if(i > 1) {
smin(f[0][i - 1][j], f[0][i][j] + (sumw[n] - (sumw[j] - sumw[i - 1])) * (tows[i].pos - tows[i - 1].pos));
smin(f[0][i - 1][j], f[1][i][j] + (sumw[n] - (sumw[j] - sumw[i - 1])) * (tows[j].pos - tows[i - 1].pos));
}
if(j < n) {
smin(f[1][i][j + 1], f[1][i][j] + (sumw[n] - (sumw[j] - sumw[i - 1])) * (tows[j + 1].pos - tows[j].pos));
smin(f[1][i][j + 1], f[0][i][j] + (sumw[n] - (sumw[j] - sumw[i - 1])) * (tows[j + 1].pos - tows[i].pos));
}
}
}
long long res = smin(f[0][1][n], f[1][1][n]);
printf(AUTO, res);
}
int main() {
freopen("power.in", "r", stdin);
freopen("power.out", "w", stdout);
init();
solve();
return 0;
}
（番外话）
时间： 2024-08-25 00:10:22