hdu 3339 In Action

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

题意：一辆坦克只能占领一个核电站，有n个核站，m条路。两个核战间可能有更短的路。每个核战都有相应的核值。问控制一半以上的核值，需要最少的油费？

思路：先别看核值，用floyd() 求出所有核战的的最短路程。 然后用 01背包求出各油费能获得的最大核值。再从小开始遍历数据，碰到核值大于一半的就输出。

#include <stdio.h>#define max_size 105#define max 10005int num[max_size][max_size], p[max_size], s[max];int main() {    int t;    scanf("%d", &t);    while(t--) {        int n, m;        scanf("%d%d", &n, &m);        for(int i = 0; i <= n; ++i)            for(int j = 0; j <= n; ++j)                if(i == j)                    num[i][j] = 0;                else                    num[i][j] = max;        for(int i = 0; i < m; ++i) {            int a, b, c;            scanf("%d%d%d", &a, &b, &c);            if(num[a][b] > c)                num[a][b] = num[b][a] = c;                }        for(int k = 0; k <= n; ++k)            for(int i = 0; i <= n; ++i)                if(num[i][k] != max)                    for(int j = 0; j <= n; ++j)                        if(num[i][j] > num[i][k] + num[k][j])                            num[i][j] = num[i][k] + num[k][j];        int dian = 0;        for(int i = 1; i <= n; ++i) {            scanf("%d", &p[i]);            dian += p[i];        }        int sum = 0, count = 0;        for(int i = 1; i <= n; ++i)            if(num[0][i] != max)                sum += num[0][i];            else                count++;        for(int i = 0; i <= sum; ++i)            s[i] = 0;        for(int i = 1; i <= n; ++i)            for(int j = sum; j >= num[0][i]; --j)                s[j] = s[j] > p[i] + s[j - num[0][i]] ? s[j] : p[i] + s[j - num[0][i]];        dian = dian / 2 + 1;        int result = 0;        for(int i = 0; i <= sum; ++i)            if(s[i] >= dian) {                result = i;                break;            }        if(result)            printf("%d\n", result);        else            printf("impossible\n");        }}