Hdu 3339 In Action

 

  
    

   
     
   
      1 //1. tank必须停在点上才能占领
 2  //2. 可以转化成0-1背包，包的“容量”是每个点的能量值，物品的value是耗油量，f[i][j]状态表示达到能量值j，需要的最小耗油量
 3  // 需要小的变形 f[i][j] = min(f[i][j], f[i][j+1], f[i-val[i]][j] + d[i]);
 4  //3. 先用dijstra求出0点到1~n个点的最短路 d[i];
 5 //4. 边的权值可能为0 
 6 //*5. 可能有多重边，输入时要处理
 7 #include <stdio.h>
 8 #include <string.h>
 9 #define Min(x,y) ((x)<(y)?(x):(y))
10 #define INF 0xfffffff
11 #define NL 110
12 
13 int n, m;
14 int e[NL][NL];
15 int d[NL];
16 int val[NL];
17 int f[NL*NL];
18 bool flg[NL];
19 
20 void Dij()
21 {
22  int i, j;
23  for (i=1; i<=n; i++)
24  d[i] = INF;
25  d[0] = 0;
26  memset(flg, 0, sizeof(flg));
27  int t = n;
28  while (t--) {
29  int min = INF, u;
30  for (i=0; i<=n; i++) {
31  if (!flg[i] && d[i] < min) {
32  min = d[i];
33  u = i;
34  }
35  }
36  flg[u] = 1;
37  for (int v=0; v<=n; v++) {
38  if (e[u][v] >= 0) {
39  if (d[v] > d[u] + e[u][v])
40  d[v] = d[u] + e[u][v];
41  }
42  }
43  }
44 }
45 
46 int main()
47 {
48  int t;
49  int i, j;
50  scanf("%d", &t);
51  while (t--) {
52  scanf("%d%d", &n, &m);
53  int a, b, c;
54  memset(e, -1, sizeof(e));
55  for (i=0; i<m; i++) {
56  scanf("%d%d%d", &a, &b, &c);
57  if (e[a][b] == -1) {
58  e[a][b] = e[b][a] = c;
59  }else {
60  e[a][b] = e[b][a] = Min(e[a][b], c);
61  }
62  }
63  int sum = 0;
64  for (i=1; i<=n; i++) {
65  scanf("%d", &val[i]);
66  sum += val[i];
67  }
68  Dij();
69  for (i=0; i<=sum+1; i++)
70  f[i] = INF;
71  f[0] = 0;
72  for (i=1; i<=n; i++) {
73  for (j=sum; j>=0; j--) {
74  if (j>=val[i]) f[j] = Min(f[j], Min(f[j+1], f[j-val[i]] + d[i]));
75  else f[j] = Min(f[j], f[j+1]);
76  }
77  }
78  for (j=sum; j>=0; j--) {
79  f[j] = Min(f[j], f[j+1]);
80  }
81  int min = f[sum/2+1];
82  if (min == INF) printf("impossible\n");
83  else printf("%d\n", min);
84  }
85  return 0;
86 }