這個專欄開始介紹一些《ACM國際大學生程序設計競賽題解》上的競賽題目，讀者可以配合zju/poj/uva的在線測評系統提交代碼(今天zoj貌似崩了)。
其實看書名也能看出來這本書的思路，就是一本題解書，簡單暴力的通過題目的堆疊來提升解決編程問題的能力。
那么下面開始探索吧。
poj1037：
Description
For years, computer scientists have been trying to find efficient solutions to different computing problems. For some of them efficient algorithms are already available, these are the "easy" problems like sorting, evaluating a polynomial or finding the shortest path in a graph. For the "hard" ones only exponential-time algorithms are known. The traveling-salesman problem belongs to this latter group. Given a set of N towns and roads between these towns, the problem is to compute the shortest path allowing a salesman to visit each of the towns once and only once and return to the starting point.
Problem
The president of Gridland has hired you to design a program that calculates the length of the shortest traveling-salesman tour for the towns in the country. In Gridland, there is one town at each of the points of a rectangular grid. Roads run from every town in the directions North, Northwest, West, Southwest, South, Southeast, East, and Northeast, provided that there is a neighbouring town in that direction. The distance between neighbouring towns in directions North-South or East-West is 1 unit. The length of the roads is measured by the Euclidean distance. For example, Figure 7 shows 2 * 3-Gridland, i.e., a rectangular grid of dimensions 2 by 3. In 2 * 3-Gridland, the shortest tour has length 6.
題目大意：給出一個nxm的矩陣，現在從某點開始遍歷所有點並回到起始點，問最少的遍歷路程是多少？(從某點出發有8個方向，行上相鄰的點之間的距離是1。)
數理分析：其實這道題目的描述非常有誤導性，多次強調所謂“旅行銷售員問題”，也就是當現在還沒有得到很好解決的著名的TSP問題，這就很容易將人帶向很混亂的思維。但是這道題目基於比較特殊的矩陣圖，我們應該能夠更加簡單的方法。
我們首先應該考慮到的一個貪心策略，行走距離為1的利益最大化就是訪問到了一個節點(當然不算起始節點)，那么如果有這樣一種方案使得我們，遍歷距離+1，都會導致多訪問了1個節點，那么這最終一定會導致最小的路程，即節點個數mn。
那么下面我們所關注的問題便是，是否存在這樣一個策略呢？如果m列是偶數，那么連接相鄰節點形成小正方格，就會形成奇數列個小方格，這使得我們能夠有一個走“S”型的鋸齒狀的遍歷策略，能夠使我們完成貪心策略。
即我們能夠得到結論，如果m、n當中有一個是偶數，最小距離都是mn。
那么如果m、n都是奇數呢？依然從m列入手，我們基於(m-1)xn這樣一個矩陣，那么我們可以采取和上面類似的思路進行貪心化的遍歷，由此我們其實能夠看到，對於這種情況，是沒有一種路程為mn的便利方案的，但對於剩下的兩列，一開始我們令遍歷路線走一個長為1.41的“斜邊”，然后由原來的縱向S型變成橫向S型，即路程為mn + 0.41，這是除去路程為mn的最優方案。
上文給出的文字描述可能過於抽象，讀者可以自行畫圖感受一下。
綜合起來我們能夠得到這道問題線性算法：
m、n存在一個偶數，結果是mn。
否自，結果是mn + 0.41.
簡單的參考代碼如下：
#include<cstdio> using namespace std; int main() { int t; scanf("%d",&t); int tt =1; while(t--) { int n , m; scanf("%d %d",&n,&m); if(n % 2 == 0 || m%2 == 0) printf("Scenario #%d:\n%d.00\n",tt++,m*n); else printf("Scenario #%d:\n%d.41\n",tt++,m*n); printf("\n"); } }
poj1003:
Description
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.
題目大意：給出一個距離dis，表示堆在桌子上的撲克的伸出距離，問你至少需要多少張撲克能夠堆出這個伸出距離。
數理分析：其實這道題目的描述就是給出了我們的解題方法，它告訴我們n張撲克能夠伸出的最遠距離是∑1/i，i∈[2,i+1]，這就使得問題變得簡單了很多。但是這個問題的原始模型其實是基於長度為2、質量為1的撲克的條件，它其實是調和數的基本模型。如果讀者有興趣看一看這個計算公式怎么得來的，可以看一下筆者的《<具體數學>——特殊的數》一文中關於調和數的介紹。
參考代碼如下。
#include<cstdio> using namespace std; int main() { double dis; double temp,j; int i; while(scanf("%lf",&dis)!=EOF && dis != 0.00) { temp = 0; for(i = 2;;i++) { j = (double)i; temp += 1/j; if(temp >= dis) break; } printf("%d card(s)\n",i-1); } }
poj1005：
Description
After doing more research, Fred has learned that the land that is being lost forms a semicircle. This semicircle is part of a circle centered at (0,0), with the line that bisects the circle being the X axis. Locations below the X axis are in the water. The semicircle has an area of 0 at the beginning of year 1. (Semicircle illustrated in the Figure.)
#include<cstdio> #include<cmath> using namespace std; const double pi = 3.1415926; int main() { double x , y , r; int t,i; scanf("%d",&t); int tt = 1; while(t--) { double temp = 0; scanf("%lf%lf",&x,&y); r = x*x + y*y; for(i = 1;;i++) { temp = (double)i*100/pi; if(temp >= r) break; } printf("Property %d: This property will begin eroding in year %d.\n",tt++,i); } printf("END OF OUTPUT."); }
tju1193：
When Issac Bernand Miller takes a trip to another country, say to France, he exchanges his US dollars for French francs. The exchange rate is a real number such that when multiplied by the number of dollars gives the number of francs. For example, if the exchange rate for US dollars to French francs is 4.81724, then 10 dollars is exchanged for 48.1724 francs. Of course, you can only get hundredth of a franc, so the actual amount you get is rounded to the nearest hundredth. (We'll round .005 up to .01.) All exchanges of money between any two countries are rounded to the nearest hundredth.
Sometimes Issac's trips take him to many countries and he exchanges money from one foreign country for that of another. When he finally arrives back home, he exchanges his money back for US dollars. This has got Issac thinking about how much if his unspent US dollars is lost (or gained!) to these exchange rartes. You'll compute how much money Issac ends up with if he exchanges it many times. You'll always start with US dollars and you'll always end with US dollars.
Input
The first 5 lines of input will be the exchange rates between 5 countries, numbered 1 through 5. Line i will five the exchange rate from country i to each of the 5 countries. Thus the jth entry of line i will give the exchange rate from the currency of country i to the currency of country j. the exchange rate form country i to itself will always be 1 and country 1 will be the US. Each of the next lines will indicate a trip and be of the form
N c1 c2 ... cn m
Where 1 ≤ n ≤ 10 and c1, ..., cn are integers from 2 through 5 indicating the order in which Issac visits the countries. (A value of n = 0 indicates end of input, in which case there will be no more numbers on the line.) So, his trip will be 1 -> c1 -> c2 -> ... -> cn -> 1. the real number m will be the amount of US dollars at the start of the trip.
Output
Each trip will generate one line of output giving the amount of US dollars upon his return home from the trip. The amount should be fiven to the nearest cent, and should be displayed in the usual form with cents given to the right of the decimal point, as shown in the sample output. If the amount is less than one dollar, the output should have a zero in the dollars place.
題目大意：一次給出序號為1、2、3、4、5這5個國家之間的貨幣匯率，然后給出周游順序<c1,c2,c3...cn>，每到一個國家，我們都將原有的外幣換成這個國家的錢，那么給出初始金額m的情況下，完成<1,c1,...cn,1>，最終手頭的錢會變成多少，結果應四舍五入至百分位。
數理分析：在理解了題意之后，能夠看到這是一道很基礎的模擬題目。需要注意的細節是，所謂四舍五入至百分位，是要求每次交換都按照這個要求，而不是得到最終的money然后四舍五入，二者在多次交換后會產生可見的誤差。第3組數據就是明顯的例子。
簡單的參考代碼如下：
#include<cstdio> using namespace std; double rate[6][6]; int country[15]; int main() { int t,n; for(int i = 1;i <= 5;i++) for(int j = 1;j <= 5;j++) scanf("%lf",&rate[i][j]); double money; while(scanf("%d",&n) && n != 0) { country[n+2] = 1; country[1] = 1; for(int i = 2;i <= n + 1;i++) scanf("%d",&country[i]); scanf("%lf",&money); for(int i = 1;i <= n + 1;i++) { money *= rate[country[i]][country[i+1]]; money = (int)(money*100 + 0.5); //每次交換都應該進行百分位的四舍五入 money /= 100; } printf("%.2lf\n",money); } }
poj1046：
Description
Input
Output
If there are more than one color with the same smallest distance, please output the color given first in the color set
#include<cstdio> using namespace std; struct color { int r ,g ,b; }a[10000]; int main() { for(int i = 1;i <= 16;i++) scanf("%d%d%d",&a[i].r,&a[i].g,&a[i].b); int rr , gg , bb; while(scanf("%d%d%d",&rr,&gg,&bb) && rr+gg+bb >= 0) { int Min = 99999999; int temp , tempi; for(int i = 1;i <= 16;i++) { temp = (a[i].r-rr)*(a[i].r-rr) + (a[i].g-gg)*(a[i].g-gg) + (a[i].b-bb)*(a[i].b-bb); if(temp < Min) { Min = temp; tempi = i; } } printf("(%d,%d,%d) maps to (%d,%d,%d)\n",rr,gg,bb,a[tempi].r,a[tempi].g,a[tempi].b); } }
zoj1078:
Statement of the Problem
We say that a number is a palindrom if it is the sane when read from left to right or from right to left. For example, the number 75457 is a palindrom.
Of course, the property depends on the basis in which is number is represented. The number 17 is not a palindrom in base 10, but its representation in base 2 (10001) is a palindrom.
The objective of this problem is to verify if a set of given numbers are palindroms in any basis from 2 to 16.
Input Format
Several integer numbers comprise the input. Each number 0 < n < 50000 is given in decimal basis in a separate line. The input ends with a zero.
Output Format
Your program must print the message Number i is palindrom in basis where I is the given number, followed by the basis where the representation of the number is a palindrom. If the number is not a palindrom in any basis between 2 and 16, your program must print the message Number i is not palindrom.
題目大意：給出一個十進制數字n，判斷n在x進制下是否會問，x∈[2,16]，並輸出x。
數理分析：這道題目分為兩個步驟，第一是進制轉化，第二是判斷回文。判斷回文根據其定義進行模擬判斷即可，下面說一下如何進行進制轉化。
我們知道，整數x表示成r進制的形式為，x = x1*r^0 + x2*r^1 + x3*r^2 +...，那么我們采取類似快速冪的過程，通過計算x%r，我們能夠得到x1，隨后我們令x = x/r，再計算x%r將得到x2......依次類推，便可得到r進制數x。
簡單的參考代碼如下。
#include<cstdio> using namespace std; int main() { int n; int i , j; while(scanf("%d",&n) && n) { int sign = 0; char c[30]; int base[17] = {0}; for(i = 2;i <= 16;i++) { int m = n; int len = 0; while(m) { c[len++] = m%i; m = m/i; } sign = 1; for(j = 0;j <len/2 &sign ;j++) if(c[j] != c[len-j-1]) sign = 0; if(sign) base[i] = 1; } sign = 1; for(i = 2;i <= 16;i++) if(base[i] == 1) sign = 0; if(sign) printf("Number %d is not a palindrom\n",n); else { printf("Number %d is palindrom in basis",n); for(i = 2;i <= 16;i++) if(base[i] == 1) printf(" %d",i); printf("\n"); } } }