2013
12-11

# Pollution

The managers of a chemical plant, which is notorious for its high pollution, plan to adopt a newly developed device in order to reduce the amount of contaminants emitted. However, engineers in the plant are against this plan, as they doubt the usefulness of the device. As engineers only believe in experimental results, managers decide to hire programmers to make a numerical experiment to convince the engineers.

The new pollution-reducing device consists of several tanks with pipes connecting the tanks. You may assume there is at most one pipe between two tanks. Two tanks are called adjacent if a pipe connects them. When operating, the contaminant circulates in the device among these tanks.

As shown in the Figure-1, the contaminant in one tank in time t, will equally distribute into all adjacent tanks in the time t+1. In other words, if we use Xit to denote the amount of contaminant in tank i at time t, we can use the following formula:

where Iij=1 if tank i and tank j are adjacent, otherwise Iij=0, and where dj is the number of tanks adjacent to tank j. If no tank is adjacent to tank i, we have Xit+1=Xit. The managers, as well as the engineers, want to know that given the initial amount of contaminant in each tank, how the contaminant will be distributed in all the tanks after a long period of time in circulation. Namely, given Xi0 for all i, what are Xit when the difference between Xit and Xit+1 is so small that it can be ignored. You may assume that this condition will ALWAYS be attained from an initial case in this problem.

The first line of the input contains one integer T (1 <= T <= 10), the number of test cases. T cases then follow. For each test case, the first line consists of two integers: N and M where(1 <= N <= 100, 0 <= M <= N*(N-1)/2), is the number of tanks and pipes. The following N lines give the initial amount of contaminant for each tank, which are nonnegative real numbers and no larger than 100. Then the next M lines give the tanks that each pipe connects, as "A B" (1 <= A, B <= N, A != B) denotes there is a pipe between tank A and tank B.

For each test case, output the final amount of contaminant Xit+1 (one per line), followed by a blank line. The number should be rounded to three digits after the decimal point.

2
3 3
1
0
0
1 2
2 3
3 1
4 4
1
0
0
1
1 2
2 3
3 1
3 4

0.333
0.333
0.333

0.500
0.500
0.750
0.250

1. #include <cstdio>
#include <algorithm>

struct LWPair{
int l,w;
};

int main() {
//freopen("input.txt","r",stdin);
const int MAXSIZE=5000, MAXVAL=10000;
LWPair sticks[MAXSIZE];
int store[MAXSIZE];
int ncase, nstick, length,width, tmp, time, i,j;
if(scanf("%d",&ncase)!=1) return -1;
while(ncase– && scanf("%d",&nstick)==1) {
for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
for(time=-1,i=0;i<nstick;++i) {
tmp=sticks .w;
for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
if(j==time) { store[++time]=tmp; }
else { store[j+1]=tmp; }
}
printf("%dn",time+1);
}
return 0;
}

2. 嗯 分析得很到位，确实用模板编程能让面试官对你的印象更好。在设置辅助栈的时候可以这样：push时，比较要push的elem和辅助栈的栈顶，elem<=min.top()，则min.push(elem).否则只要push（elem）就好。在pop的时候，比较stack.top()与min.top(),if(stack.top()<=min.top()),则{stack.pop();min.pop();}，否则{stack.pop();}.

3. 算法是程序的灵魂，算法分简单和复杂，如果不搞大数据类，程序员了解一下简单点的算法也是可以的，但是会算法的一定要会编程才行，程序员不一定要会算法，利于自己项目需要的可以简单了解。