UVA10123

题目的意思就是要把木块拿下来，但天平不能倒，天平的支点在-1.5 和 1.5的位置，如果 左边支点 左边的力矩大于右边，就有左倒， 右边支点 右边的力矩大于左边就会右倒，

但是大的量要超过木板质量。

首先要把问题看成是放木块。

先用两个数组来把左右两边的木块存起来，然后按力矩升序排列，然后按顺序一个个放，不会倒就放上去，会倒就放另一边的，知道放完。

#include<iostream>#include<math.h>#include<algorithm>using namespace std;const int N = 20 + 5;struct wood {int pos;int weight;int torque;}leftw[N],rightw[N],res[N];bool ok ;int num,l,wei;int leftnum ,rightnum,left,right,midnum;int cmp (wood a ,wood b) {return a.torque < b.torque;}bool yes(int l ,int r) {int suml = 0;int sumr = 0;for (int i = 0 ; i < l ;i++) {suml += (-leftw[i].pos * 2 - 3) * leftw[i].weight;}for (int i = 0 ; i < r ;i++) {sumr += (rightw[i].pos * 2 + 3) * rightw[i].weight;}sumr += wei;if (suml > sumr)return false;suml = 0;   sumr = 0;for (int i = 0; i < l; i++) {suml += (-leftw[i].pos * 2 + 3) * leftw[i].weight;}for (int i = 0; i < r; i++) {sumr += (rightw[i].pos * 2 - 3) * rightw[i].weight;}suml += wei;if (sumr > suml)   return false;return true;;}void dfs (int cur ,int l ,int r) {if (cur == num) {for (int i = cur - 1 ; i >= 0 ;i--) {cout <<res[i].pos <<" "<<res[i].weight<<endl;}ok = true;return;}for (int i = l ; i < leftnum ;i++) {if(yes(i,r) && !ok) {res[cur].pos = leftw[i].pos;res[cur].weight = leftw[i].weight;dfs(cur + 1, i + 1,r);}elsebreak;}for (int i = r ; i < rightnum ;i++) {if(yes(l,i) && !ok) {res[cur].pos = rightw[i].pos;res[cur].weight = rightw[i].weight;dfs(cur + 1, l, i + 1);}elsebreak;}}int main () {int a,b;int T = 1;while (cin >> l >> wei >> num && l + wei +num) {leftnum = rightnum = 0 ;for (int i = 0 ; i < num ; i++) {cin >> a >> b;if (a >= 0) {rightw[rightnum].pos = a;rightw[rightnum].weight = b;rightw[rightnum++].torque = (a * 2 - 3) * b;}if (a < 0) {leftw[leftnum].pos = a;leftw[leftnum].weight = b;leftw[leftnum++].torque = (-a * 2 - 3) * b;}}sort(leftw ,leftw + leftnum ,cmp);sort(rightw , rightw + rightnum ,cmp);wei = wei * 3;ok = false;cout << "Case "<< T++ << ":\n";dfs(0,0,0);if(!ok)cout << "Impossible" <<endl; }return 0;}