In Action
Time Limit: 2000/1000 MS
(Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 3869 Accepted
Submission(s): 1237
Problem Description
Since
1945, when the first nuclear bomb was exploded by the Manhattan Project team in
the US, the number of nuclear weapons have soared across the
globe.
Nowadays,the crazy boy in FZU named AekdyCoin possesses some nuclear
weapons and wanna destroy our world. Fortunately, our mysterious spy-net has
gotten his plan. Now, we need to stop it.
But the arduous task is obviously
not easy. First of all, we know that the operating system of the nuclear weapon
consists of some connected electric stations, which forms a huge and complex
electric network. Every electric station has its power value. To start the
nuclear weapon, it must cost half of the electric network‘s power. So first of
all, we need to make more than half of the power diasbled. Our tanks are ready
for our action in the base(ID is 0), and we must drive them on the road. As for
a electric station, we control them if and only if our tanks stop there. 1 unit
distance costs 1 unit oil. And we have enough tanks to use.
Now our commander
wants to know the minimal oil cost in this action.
Input
The first line of the input contains a single integer
T, specifying the number of testcase in the file.
For each case, first line
is the integer n(1<= n<= 100), m(1<= m<= 10000), specifying the
number of the stations(the IDs are 1,2,3...n), and the number of the roads
between the station(bi-direction).
Then m lines follow, each line is interger
st(0<= st<= n), ed(0<= ed<= n), dis(0<= dis<= 100), specifying
the start point, end point, and the distance between.
Then n lines follow,
each line is a interger pow(1<= pow<= 100), specifying the electric
station‘s power by ID order.
Output
The minimal oil cost in this action.
If not exist
print "impossible"(without quotes).
Sample Input
2
2 3
0 2 9
2 1 3
1 0 2
1
3
2 1
2 1 3
1
3
Sample Output
5
impossible
Author
[email protected]
Source
HDOJ
Monthly Contest – 2010.03.06
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1 //187MS 616K 1856 B C++
2 /*
3
4 题意：
5 题意挺重要的。一个源点0，源点每到一个站可得到一个pow[i]，求得到总pow值一半以上时最短行走距离。
6
7 最短路+01背包：
8 首先最短路求出源点0到每一个点的距离，然后得出n个点，每个点有一个距离花费d，和一个价值权值w，
9 以到所有的点最短路程和作为背包容量，距离作为花费，权值作为价值，01背包。
10 当dp[j]>sum_pow/2+1时的距离和j为所求。
11
12 */
13 #include<iostream>
14 #include<queue>
15 #include<vector>
16 #define N 105
17 #define inf 0x7ffffff
18 using namespace std;
19 typedef struct node{
20 int d,w;
21 }node;
22 node p[N];
23 int vis[N];
24 int d[N];
25 int dp[N*N];
26 vector<node>V[N];
27 int n;
28 int Max(int x,int y)
29 {
30 return x>y?x:y;
31 }
32 void spfa(int s)
33 {
34 for(int i=0;i<=n;i++)
35 d[i]=inf;
36 d[s]=0;
37 vis[s]=1;
38 queue<int>Q;
39 Q.push(s);
40 while(!Q.empty()){
41 int u=Q.front();
42 Q.pop();
43 vis[u]=0;
44 int m=V[u].size();
45 for(int i=0;i<m;i++){
46 int v=V[u][i].d;
47 int w=V[u][i].w;
48 if(d[v]>d[u]+w){
49 d[v]=d[u]+w;
50 if(!vis[v]){
51 vis[v]=1;
52 Q.push(v);
53 }
54 }
55 }
56 }
57 }
58 int main(void)
59 {
60 int t,m;
61 int u,v,w,pow;
62 scanf("%d",&t);
63 while(t--)
64 {
65 int sd=0,sw=0;
66 scanf("%d%d",&n,&m);
67 for(int i=0;i<=n;i++) V[i].clear();
68 for(int i=0;i<m;i++){
69 scanf("%d%d%d",&u,&v,&w);
70 node q={v,w};
71 V[u].push_back(q);
72 q.d=u;
73 V[v].push_back(q);
74 }
75 for(int i=1;i<=n;i++){
76 scanf("%d",&p[i].w);
77 sw+=p[i].w;
78 }
79 spfa(0);
80 for(int i=1;i<=n;i++){
81 p[i].d=d[i];
82 if(d[i]!=inf)
83 sd+=p[i].d;
84 }
85 memset(dp,0,sizeof(dp));
86 for(int i=1;i<=n;i++)
87 for(int j=sd;j>=p[i].d;j--)
88 dp[j]=Max(dp[j],dp[j-p[i].d]+p[i].w);
89 int ans=0;
90 for(int i=1;i<=sd;i++)
91 if(dp[i]>=sw/2+1){
92 ans=i;break;
93 }
94 if(ans==0) puts("impossible");
95 else printf("%d\n",ans);
96 }
97 return 0;
98 }
hdu 3339 In Action (最短路径+01背包),布布扣,bubuko.com