HDU3339 In Action（最短路+01背包）
In Action
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4189 Accepted Submission(s): 1341
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 wZ喎?http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">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"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
第一次看見這個題目時時和ZMC作比賽，結果他AK了，我這個題看不懂，就這麼過去了，復習最短路時又遇見了這次不能再錯過了吧，又重新的看的題；
這是一個最短路+01背包的問題.
大體題意：是n個發電站，m條路，每條路有各自的距離，每個發電站有各自的發電量，現在需要炸毀它們，一輛坦克只能炸毀一個發電站，而且需要炸毀的發電廠的發電量需要大於所有發電站所產生的總電量的一半，求坦克走的最短距離。
因為題目說明不超過100個點，所以可以用弗洛伊德算法，把所有的0到各個發電站的距離找出來，然後把0能到的各個發電站的距離相加作為背包的容量，以dp數組儲存的值作為炸毀的發電站的電量，最後進行判斷，如果dp[i]的值大於所有發電站的總電量，輸出i的值，此時i的值就是坦克炸毀發電站使城市供電癱瘓所能走的最短距離
#include
#include
#include
#define INF 0x3f3f3f3f
#define N 305
int map[N][N];
int n,m;
int a[10010];
int dp[100001];
void floyd()
{
for(int k=0;k<=n;k++)
{
for(int i=0;i<=n;i++)
{
for(int j=0;j<=n;j++)
{
if(map[i][j] > map[i][k]+map[k][j])
{
map[i][j] = map[i][k] + map[k][j];
}
}
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=0; i<=n; i++)
{
for(int j=0; j<=n; j++)
{
map[i][j] = INF;
}
map[i][i] = 0;
}
int x,y,z;
for(int i=0; i<m; i++)
{
scanf("%d%d%d",&x,&y,&z);
if(map[x][y]>z)
{
map[x][y] = z;
map[y][x] = z;
}
}
floyd();
int sum = 0;
int count = 0;
for(int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
sum += a[i];
if(map[0][i] != INF)
{
count += map[0][i];
}
}
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
for(int j=count;j>=map[0][i];j--)
{
if(dp[j-map[0][i]] + a[i] > dp[j])
{
dp[j] = dp[j-map[0][i]] + a[i];
}
}
}
int flag = 0;
for(int i=1;i<=count;i++)
{
if(dp[i] > sum/2.0)
{
flag = 1;
printf("%d\n",i);
break;
}
}
if(flag == 0)
{
printf("impossible\n");
}
}
return 0;
}
