Mathmen
Time Limit: 1000MS Memory limit: 65536K
题目描述
Mathmen
love mathematics, and they live on the number line. All the mathmen
spend all their time on solving mathematical problems and proving theorems.
Like humen beings, mathmen don‘t live in the same country there are
n mathmen countries on difierent positions of the number line (country i
lives at position xi (one point on the number line), and xi < xi + 1 for all
1 <= i <= n - 1). Mathmen in difierent countries are good at difierent
areas of mathematics. For example, mathmen living in country 1 are good
at geometry, while mathmen living in country 2 are experts on algebra.
Lately,
all the mathmen nations have collaborated on a huge problem, which involves
all areas of mathematics, and each mathmen country is responsible for
solving the part they are good at. After hours of work (they are really good
at mathematics, and no problem would cost them more than one day to
solve), they all finish their jobs. Now, they need to collect the result in every
mathmen country, and combine them together to create the solution to the
huge problem. So they decide to let one mathman collect all the results and
send it to one country, where all the work will be merged.
As
mathmen are devoted to solving mathematical problems, their legs has
become very weak after millions of years of evolution. Therefore, they have
to ride mathships to travel on the number line. Their are M types of mathships,
each of which has a limit of traveling distance and a cost of IQ (yes,
you need to be very brave to take mathships, you would never be able to
get back the lost IQ).
There
are two seasons in the mathmen world Positive and Negative. Now it
is Positive, so all the mathships travels on the number line from left to right.
Therefore, one man from country 1 must be selected as the volunteer to
collect the results in every country and send to to country n.
There
is at least one mathship of each type in every country. So, after picking
the results from country 1, the volunteer needs to select one mathship with
a limit of traveling distance at least the distance between country 1 and country
2 (the distance is x2 - x1), and ride it from country 1 to country 2.
Meahwhile,
this trip will cost him the corresponding amount of IQ. Then, he picks
the result from country 2, choose another suitable mathship (the old mathship
will be maintained in country 2, and can not continue to be used), and
ride it to country 3, and lose some IQ. He will repeat this process, until he
reaches country n.
Mathmen
care about their IQ a lot, so the volunteer wants to minimize the
total cost of his IQ. Could you please write program to help him select the right mathships to take?
输入
The input contains multiple test cases. The first line of the input is an integer C,
indicating the number of test cases in the input.
Each
test case begins with a line containing two integers, n (2 <= n <= 10000)
and m (1 <= m <= 100000), which is the number of mathmen countries
and the number of mathship types. The next line contains n integers, ranging
from -10^9to10^9(inclusive), indicating the positions of the mathmen countries.
Then, m lines follows, each containing two non-negative integers no
more than 2*109, which are the limit of traveling distance and the cost of
IQ.
输出
For each test case, output one line containing the minimum cost of IQ. If the volunteer
can never reach country n, output "Impossible" (without quotation marks).
示例输入
2 3 3 0 3 10 1 0 6 1 10 10 2 1 0 1000 100 0
示例输出
11 Impossible
题目大意是给出n个城市每个城市的坐标递增，每个城市中有m个交通工具，每个交通工具有可以行进的距离和耗油，不满最大距离时，按最大距离算，问能不能从0号城市走到n-1号城市。
对每个城市之间的距离进行由大到小排序，然后对没见交通工具可以走的距离进行排序，然后用优先队列找出最小的值。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std ;
#define LL long long
priority_queue <LL,vector<LL>,greater<LL> > que ;
struct node{
LL x , y ;
}p[110000];
LL a[11000] , dis[11000] ;
int cmp(node a,node b) {
return a.x < b.x ;
}
int main() {
int t , n , m ;
LL ans ;
int i , j ;
scanf("%d", &t) ;
while( t-- ) {
ans = 0 ;
scanf("%d %d", &n, &m) ;
for(i = 0 ; i < n ; i++) {
scanf("%lld", &a[i]) ;
if( i ) dis[i] = a[i] - a[i-1] ;
}
for(i = 0 ; i < m ; i++)
scanf("%lld %lld", &p[i].x, &p[i].y) ;
sort(dis+1,dis+n) ;
sort(p,p+m,cmp) ;
while( !que.empty() ) que.pop() ;
j = m-1 ;
for(i = n-1 ; i >= 1 ; i--) {
while( j >= 0 && p[j].x >= dis[i] ) {
que.push(p[j].y) ;
j-- ;
}
if( que.empty() ) break ;
ans += que.top() ;
}
if( i >= 1 )
printf("Impossible\n") ;
else
printf("%lld\n", ans) ;
}
return 0 ;
}