标签：div process nim cli client ble acm ogr memset
Your clients desire to buy a certain number (M) of bricks, which for, uh, religious reasons must be of different types. They will be melted together, and the resultant mixture must have a concentration of at least CMin and at most CMax grams of copper per kilogram. Their goal is to pick the M types of brick so that the mixture has the correct concentration and the price of the collection is minimized. You must figure out how to do this. M, CMin, and CMax will vary depending on the client.
The second part consists of a line containing a number C (1 ≤ C ≤ 100), the number of clients you are serving, followed by C lines containing M (1 ≤ M ≤ 20), CMin (1 ≤ CMin ≤ 999), and CMax (1 ≤ CMax ≤ 999) for each client.
All input numbers will be positive integers.
11 550 300 550 200 700 340 300 140 600 780 930 785 730 280 678 420 999 900 485 390 888 800 3 2 500 620 9 550 590 9 610 620
420 impossible 3635
Source: Waterloo
Local Contest Jun. 19, 1999
dp过程：
dp[0][0] = 0;//dp[i][j]表示i块砖铜总量j最小价格 d = min(20,n); for(k=1;k<=n;k++){ for(j=20000;j>=con[k];j--){ for(i=1;i<=d;i++){ if(dp[i-1][j-con[k]]!=INF){ dp[i][j] = min(dp[i][j],dp[i-1][j-con[k]]+prc[k]); } } } }
开始时先循环i后循环j，结果导致每个铜砖被重复使用。
全部代码：
#include <stdio.h> #include <algorithm> #include <math.h> #include <memory.h> using namespace std; const int INF=0x3f3f3f3f; int con[202],prc[202],dp[22][20002]; int n,c,m,cmin,cmax; int main(){ int i,j,k,d; while(~scanf("%d",&n)){ memset(dp,INF,sizeof(dp)); for(i=1;i<=n;i++){ scanf("%d%d",&con[i],&prc[i]); } dp[0][0] = 0;//dp[i][j]表示i块砖铜总量j最小价格 d = min(20,n); for(k=1;k<=n;k++){ for(j=20000;j>=con[k];j--){ for(i=1;i<=d;i++){ if(dp[i-1][j-con[k]]!=INF){ dp[i][j] = min(dp[i][j],dp[i-1][j-con[k]]+prc[k]); } } } } /*for(i=0;i<=d;i++){ for(j=0;j<=20000;j++){ if(dp[i][j]==INF) continue; else printf("%d %d %d ",i,j,dp[i][j]); } printf("\n"); }*/ scanf("%d",&c); while(c--){ scanf("%d%d%d",&m,&cmin,&cmax); if(m>n) printf("impossible\n"); else { int ans = INF; int l = m*cmin; int h = m*cmax; for(i=l;i<=h;i++){ if(dp[m][i]<ans){ ans = dp[m][i]; } } if(ans!=INF) printf("%d\n",ans); else printf("impossible\n"); } } } return 0; }
标签：div process nim cli client ble acm ogr memset
原文地址：http://www.cnblogs.com/shenchuguimo/p/6358757.html
