HDU3339 - In Action - 最短路+01背包dp

1.题目描述：

In Action

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5964    Accepted Submission(s): 2007

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

Lost@HDU

 

Source

HDOJ Monthly Contest – 2010.03.06

 

Recommend

lcy

 

2.题意概述：

给出n,m，代表有n个发电站，m条路，然后给出每条路和该路消耗的油，然后是n行，每行是相应的发电站的能量，要注意，每个发电站必须停一辆坦克才能控制这个发电站，只有控制的总能量超过一半能使其瘫痪

3.解题思路：

因为是求最少油耗，所以是最短路，但是在此路上的各个站都有相应的能量，那么就是看在大于一半的情况下，去不去这个站并取最小油耗的问题，这就是01背包取和不取的问题了 

4.AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define N 111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
struct node
{
	int to, val;
	node(int a, int b) { to = a; val = b; }
};
vector<node> mp[N];
int dis[N], power[N], dp[maxn];
bool vis[N];
void spfa(int sta, int n)
{
	memset(vis, 0, sizeof(vis));
	fill(dis, dis + n + 1, INF);
	deque<int> q;
	vis[sta] = 1;
	dis[sta] = 0;
	q.push_back(sta);
	while (!q.empty())
	{
		int u = q.front();
		q.pop_front();
		vis[u] = 0;
		int sz = mp[u].size();
		for (int i = 0; i < sz; i++)
		{
			int v = mp[u][i].to;
			int w = mp[u][i].val;
			if (dis[v] > dis[u] + w)
			{
				dis[v] = dis[u] + w;
				if (!vis[v])
				{
					vis[v] = 1;
					if (!q.empty() && dis[v] <= dis[q.front()])
						q.push_front(v);
					else
						q.push_back(v);
				}
			}
		}
	}
}
int main()
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
	long _begin_time = clock();
#endif
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n, m;
		scanf("%d%d", &n, &m);
		for (int i = 0; i <= n; i++)
			mp[i].clear();
		memset(dp, 0, sizeof(dp));
		for (int i = 0; i < m; i++)
		{
			int u, v, w;
			scanf("%d%d%d", &u, &v, &w);
			mp[u].push_back(node(v, w));
			mp[v].push_back(node(u, w));
		}
		spfa(0, n);
		int cap = 0;
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &power[i]);
			cap += power[i];
		}
		cap /= 2;
		int way = 0;
		for (int i = 0; i <= n; i++)
			if (dis[i] != INF)
				way += dis[i];
		for (int i = 1; i <= n; i++)
		{
			if (dis[i] == INF)
				continue;
			for (int j = way; j >= dis[i]; j--)
				dp[j] = max(dp[j], dp[j - dis[i]] + power[i]);
		}
		int flag = -1;
		for (int i = 1; i <= way; i++)
			if (dp[i] > cap)
			{
				flag = i;
				break;
			}
		if (flag == -1)
			puts("impossible");
		else
			printf("%d\n", flag);
	}
#ifndef ONLINE_JUDGE
	long _end_time = clock();
	printf("time = %ld ms.", _end_time - _begin_time);
#endif
	return 0;
}