In Action

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4764    Accepted Submission(s): 1569

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

题解：01背包+最短路；

代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<algorithm>
 6 using namespace std;
 7 #define mem(x,y) memset(x,y,sizeof(x))
 8 const int INF=0x3f3f3f3f;
 9 const int MAXN=110;
10 const int MAXM=10010<<1;
11 struct Node{
12     int u,v,w;
13 };
14 Node dt[MAXM];
15 int edgnum,n;
16 int dis[MAXN];
17 int bag[10010];
18 int power[MAXN];
19 int add(int u,int v,int w){
20     dt[edgnum].u=u;dt[edgnum].v=v;dt[edgnum++].w=w;
21 }
22 void Bellman(){
23     mem(dis,INF);dis[0]=0;
24     for(int j=0;j<=n;j++){
25     for(int i=0;i<edgnum;i++){
26         int u=dt[i].u,v=dt[i].v,w=dt[i].w;
27         dis[v]=min(dis[v],dis[u]+w);
28         }
29     }
30 }
31 int main(){
32     int m,T;
33     scanf("%d",&T);
34     while(T--){
35         edgnum=0;
36         scanf("%d%d",&n,&m);
37         int u,v,w;
38         while(m--){
39             scanf("%d%d%d",&u,&v,&w);
40             add(u,v,w);add(v,u,w);
41         }
42         mem(bag,0);
43         int V=0,sum=0;
44         for(int i=1;i<=n;i++)scanf("%d",power+i),V+=power[i];
45         int flot=0;
46         Bellman();
47         for(int i=1;i<=n;i++){
48             if(dis[i]==INF){
49                 continue;
50             }
51             sum+=dis[i];
52         }
53         for(int i=1;i<=n;i++){
54             if(dis[i]!=INF)
55             for(int j=sum;j>=dis[i];j--){
56                 bag[j]=max(bag[j],bag[j-dis[i]]+power[i]);
57             }
58         }
59         int i;
60         for(i=0;i<sum;i++)if(bag[i]>V/2){//不能是（v+1）/2。。。
61             flot=1;
62             break;
63         }
64         if(flot)printf("%d\n",i);
65         else puts("impossible");
66     }
67     return 0;
68 }