题目链接：

题目

In Action

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 32768/32768 K (Java/Others)

问题描述

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.

输入

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.

输出

The minimal oil cost in this action.

If not exist print "impossible"(without quotes).

样例

input

2

2 3

0 2 9

2 1 3

1 0 2

1

3

2 1

2 1 3

1

3

output

5

impossible

题意

给你一无向图，每个节点有电力值，现在你从0点派出坦克（每单位的距离耗一单位油），每个坦克只有停在一个节点上才能摧毁电力值

问毁坏超过一半的电力的最小耗油量。

题解

跑0到所有点的最短路，然后01背包。

代码

#include<queue>

#include<vector>

#include<cstdio>

#include<cstring>

#include<iostream>

using namespace std;

const int maxn = 111;

const int maxm = 10000 + 10;

const int INF = 0x3f3f3f3f;

int mat[maxn][maxn];

int power[maxn];

int dp[maxn][maxm];

int n, m;

void init() {

	for (int i = 0; i < maxn; i++) for (int j = 0; j < maxn; j++) {

		mat[i][j] = INF;

		if (i == j) mat[i][j] = 0;

	}

	memset(power, 0, sizeof(power));

	memset(dp, 0x3f, sizeof(dp));

	for (int i = 0; i < maxn; i++) dp[i][0] = 0;

}

int main() {

	int tc;

	scanf("%d", &tc);

	while (tc--) {

		scanf("%d%d", &n, &m);

		init();

		while (m--) {

			int u, v, w;

			scanf("%d%d%d", &u, &v, &w);

			if (mat[u][v] > w)

				mat[u][v] = mat[v][u] = w;

		}

		int sum = 0;

		for (int i = 1; i <= n; i++) {

			scanf("%d", power + i);

			sum += power[i];

		}

		for (int k = 0; k <= n; k++) {

			for (int i = 0; i <= n; i++) {

				for (int j = 0; j <= n; j++) {

					if (mat[i][j] > mat[i][k] + mat[k][j])

						mat[i][j] = mat[i][k] + mat[k][j];

				}

			}

		}

		for (int i = 1; i <= n; i++) {

			for (int j = power[i]; j < maxm; j++) {

				dp[i][j] = min(dp[i][j], dp[i - 1][j]);

				dp[i][j] = min(dp[i][j], dp[i - 1][j - power[i]] + mat[0][i]);

			}

		}

		int ans = INF;

		for (int i = sum / 2 + 1; i <= sum; i++) {

			ans = min(ans, dp[n][i]);

		}

		if(ans<INF) printf("%d\n", ans);

		else printf("impossible\n");

	}

	return 0;

}