In Action
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5472 Accepted Submission(s): 1843
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.
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).
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
Lost@HDU
Source
题意：
有n+1个点，m条路，0点是起点，除0点外每个点有一个权值，经过每条路会有相应的油费，若干辆车从0点出发去占领点，要占领一半以上的总权值才行，问最少的油费。
代码：
//读错题了以为是只有一辆车。算出0点到每个点的最短路，以总路程为容量01背包找出取哪些点权值最大或以总权值为容量01背包出最小路程就行了，01背包又忘了。。。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int MAX=;
int mp[][],dis[],vis[],dp[],fei[];//dp别忘了开大
void dijk(int n)
{
for(int i=;i<=n;i++)
{
dis[i]=mp[][i];
vis[i]=;
}
vis[]=;
for(int i=;i<=n;i++)
{
int Min=MAX,sta=;
for(int j=;j<=n;j++)
{
if(!vis[j]&&dis[j]<Min)
{
Min=dis[j];
sta=j;
}
}
vis[sta]=;
for(int j=;j<=n;j++)
{
if(!vis[j]&&mp[sta][j]!=MAX&&dis[j]>dis[sta]+mp[sta][j])
dis[j]=dis[sta]+mp[sta][j];
}
}
}
int main()
{
int t,n,m,a,b,c;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
mp[i][j]=i==j?:MAX;
for(int i=;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
mp[a][b]=mp[b][a]=min(mp[a][b],c);
}
int sum=;
for(int i=;i<=n;i++)
{scanf("%d",&fei[i]);sum+=fei[i];}
dijk(n);
int V=;
for(int i=;i<=n;i++)
{
if(dis[i]!=MAX) //去掉这样的点
V+=dis[i];
}
memset(dp,,sizeof(dp));
for(int i=;i<=n;i++)
{
if(dis[i]==MAX) continue;
for(int j=V;j>=dis[i];j--)
dp[j]=max(dp[j],dp[j-dis[i]]+fei[i]);
}
int flag=;
for(int i=;i<=V;i++)
{
if(dp[i]>=sum/+)
{
printf("%d\n",i);
flag=;
break;
}
}
if(!flag) printf("impossible\n");
}
return ;
}