hdu 4568（状态压缩dp）

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 }
View Code

 

posted @ 2014-05-16 02:06  张小豪  阅读(310)  评论(0)  编辑  收藏  举报