题意：一张n*m的网格内每个点有话费，还有若干个宝藏，问一个人要走进去拿走所有宝藏在走出来的最小花费。
思路：看宝藏只有13个直接想到了状压dp[i][j]拿了哪几个前一个为j的最小花费，先bfs+优先队列预处理出最短路，然后记忆化搜索就可。
代码如下：
1 /**************************************************
2 * Author : xiaohao Z
3 * Blog : http://www.cnblogs.com/shu-xiaohao/
4 * Last modified : 2014-05-15 16:59
5 * Filename : C.cpp
6 * Description :
7 * ************************************************/
8
9 #include <iostream>
10 #include <cstdio>
11 #include <cstring>
12 #include <cstdlib>
13 #include <cmath>
14 #include <algorithm>
15 #include <queue>
16 #include <stack>
17 #include <vector>
18 #include <set>
19 #include <map>
20 #define MP(a, b) make_pair(a, b)
21 #define PB(a) push_back(a)
22
23 using namespace std;
24 typedef long long ll;
25 typedef pair<int, int> pii;
26 typedef pair<unsigned int,unsigned int> puu;
27 typedef pair<int, double> pid;
28 typedef pair<ll, int> pli;
29 typedef pair<int, ll> pil;
30
31 const int INF = 0x3f3f3f3f;
32 const double eps = 1E-6;
33 const int LEN = 210;
34 int Map[LEN][LEN], mp[LEN][LEN], out[LEN], in[LEN], dp[1<<13][20];
35 int n, m, vn, N;
36 int xx[] = {0, 0, 1,-1};
37 int yy[] = {1,-1, 0, 0};
38 pii vex[LEN];
39
40 bool J(int x, int y){
41 if(x == 1 || x == n || y == 1 || y == m) return true;
42 return false;
43 }
44
45 bool Ja(int x, int y){
46 if(x>0 && x<=n && y>0 && y<=m) return true;
47 return false;
48 }
49
50 struct P{
51 int x, y, val;
52 bool operator<(const P a) const
53 {
54 return this->val > a.val;
55 }
56 };
57
58 P mmp(int a, int b, int c){
59 P ret;
60 ret.x = a;ret.y = b;ret.val = c;
61 return ret;
62 }
63
64 int Bfs(int sx, int sy, int ex, int ey){
65 int vis[210][210] = {0};
66 priority_queue<P> q;
67 q.push(mmp(sx, sy, 0));
68 vis[sx][sy] = 1;
69 while(!q.empty()){
70 P nv = q.top();q.pop();
71 int x = nv.x, y = nv.y, val = nv.val;
72 if(ex < 0 && ey < 0){
73 if(J(x, y)) return val;
74 }
75 else if(x == ex && y == ey) return val;
76 for(int i=0; i<4; i++){
77 int tx = x + xx[i];
78 int ty = y + yy[i];
79 if(Ja(tx, ty) && !vis[tx][ty]){
80 vis[tx][ty] = 1;
81 q.push(mmp(tx, ty, val + mp[tx][ty]));
82 }
83 }
84 }
85 return -1;
86 }
87
88 void init(){
89 for(int i=0; i<vn; i++){
90 for(int j=0; j<vn; j++){
91 Map[i][j] = Bfs(vex[i].first, vex[i].second, vex[j].first, vex[j].second);
92 }
93 }
94 for(int i=0; i<vn; i++){
95 int x = vex[i].first, y = vex[i].second;
96 out[i] = Bfs(x, y, -1, -1);
97 in[i] = out[i] + mp[x][y];
98 }
99 }
100
101 int dfs(int x, int lv){
102 if(dp[x][lv] != INF) return dp[x][lv];
103 int cnt = 0;
104 for(int i=0; i<vn; i++){
105 if(x & (1 << i)) cnt ++ ;
106 }
107 if(cnt == 1) return dp[x][lv] = in[lv];
108 int ret = INF;
109 for(int i=0; i<vn; i++){
110 if((x & (1<<i)) && i != lv){
111 int &val = Map[i][lv];
112 ret = min(ret, dfs(x^(1<<lv), i) + val);
113 }
114 }
115 return dp[x][lv] = ret;
116 }
117
118 int main()
119 {
120 // freopen("in.txt", "r", stdin);
121
122 int T, a, b;
123 scanf("%d", &T);
124 while(T--){
125 scanf("%d%d", &n, &m);
126 memset(Map, 0x3f, sizeof Map);
127 memset(mp, 0, sizeof mp);
128 memset(dp, 0x3f, sizeof dp);
129 for(int i=1; i<=n; i++){
130 for(int j=1; j<=m; j++){
131 scanf("%d", &mp[i][j]);
132 if(mp[i][j] == -1) mp[i][j] = INF;
133 }
134 }
135 scanf("%d", &vn);
136 for(int i=0; i<vn; i++){
137 scanf("%d%d", &a, &b);
138 a++, b++;
139 vex[i] = MP(a, b);
140 }
141 init();
142 int ans = INF;
143 for(int i=0; i<vn; i++){
144 ans = min(ans, dfs((1<<vn)-1, i) + out[i]);
145 }
146 if(ans != INF) printf("%d\n", ans);
147 else printf("0\n");
148 }
149 return 0;
150 }
hdu 4568（状态压缩dp）,布布扣,bubuko.com
时间： 2024-08-24 23:59:58