error 1062

Text ReverseProblem Descriptionignatius likes to the write words in reverse. Given a single line of text which are written by Ignatius, you should reverse all the words and then output them.Inputthe input contains several test cases. The first line of the input was a single integer T which is the number of test cases. T test Cases follow.Each test case contains a, with several words. There'll is at the most of characters in a line.Outputfor Each test case, you should the output of the text which

between the two levels exceeds the limit mvist[j]=true;//Item J is forced to be defined as "visited" and is not involved in subsequent operations ElseVist[j]=false;//Otherwise, item J is defined as "not visited" and participates in subsequent operations} temp_price=dijkstra ();//Record the minimum distance (minimum price) of the target point 1 under level lv[i] constraints after the current trade if(Minprice>temp_price)//find the minimum price after each transaction, and final

/* The topic meaning is very simple, but should pay attention to the space idea: 1, the traversal is the character output, so need counter count (representing the starting position of the need to transpose), I used to indicate the end position of the transpose 2, for the space to special treatment; */# include HDU 1062 Text Reverse

) - if((L[j] >= min) (l[j] Len[step]; - for(j =1; J ) - if((!flag[j]) (Len[j] minl)) { -Minl =Len[j]; inTMP =J; - } toFLAG[TMP] =1; +Step =tmp; - } the returnlen[1]; * } $ Panax Notoginseng intMain () { - intI, J, T, V, MAXL, TMP; theMAXL =maxint; +scanf"%d%d", m, n); A for(i =0; I ) the for(j =0; J ) +MAP[I][J] =Maxint; - for(i =1; I ){ $scanf" %d%d%d", p[i], l[i], x[i]); $map[0][i] =P[i]; - for(j =1; J ){ -scanf"%d%d", t, v); theMap[t][i] =v;

,inty2) { $ returnSUM (k,x2,y2) +sum (k,x1-1, y1-1)-sum (k,x2,y1-1)-sum (k,x1-1, y2);Panax Notoginseng } - intMain () { theN=read (); Len=read ();len2=len1;len4=len21; + while(n--){ A intopt=read (); the if(opt==1){ + intT=read (), C=read (), L=read (), R=read (), d=read (); -x[c]= (t-d*l+len2)%Len2; $y[c]=r-l; $Add0, X[c],y[c]+x[c],1); -Add1, X[c],y[c]-x[c]+len2,1); -}Else the if(opt==2) - {Wuyi intT=read (), L=read (), r=read ();

() { * $Cin>>level_gap>>num;Panax Notoginseng for(intI=0; i){ - for(intj=0; j){ themap[i][j]=Int_max; + } A } thedis[0]=0; +min_cost=Int_max; - for(intI=1; i){ $ intValue,lev,rep_num; $Cin>>value>>lev>>Rep_num; -map[0][i]=value; -level[i]=Lev; the for(intj=0; j){ - intN;WuyiCin>>n>>value; themap[n][i]=value; - } Wu } - for(inti=level[1]-level_gap;i1];i++){ About for(intj=1; j){ $ if(level[j]>=ilev

letters name1_i, name2_i, respectively, a small hi asked two names. For 100% of the data, the nOutput For each set of test data, for each small hi query, the output line, indicating the results of the inquiry: if according to the known information, you can determine that the two people in the inquiry have a common ancestor, then output all of their common ancestor generational the lowest name of a person, otherwise output-1. Sample input 11JiaYan Jiadaihuajiada

daughter.InputThe input first line is two integers m,n (1 OutputOutput the minimum number of coins required.Sample Input1 410000 3 22 80003 50001000 2 14 2003000 2 14 20050 2 0Sample Output5250I think out of the idea and the problem realized when the special excitement.Note this question two points:1. Take the part of the offer as a path and then Dijkstra, and compare the value of each point + the shortest path length of that point.2. Record the front node of each point when it enters set, in o

,MINP; for(intcas=0; casCAs) {Minn=INF; MINP=-1; for(intI=0; ii)if(!vis[i] minn>Lowcost[i]) {Minn=Lowcost[i]; MINP=i; } if(minp==-1) return; VIS[MINP]=1; for(intI=0; ii)if(!vis[i] cost[minp][i]!=-1 (i==0|| (l[i]-minl[minp]Cost[minp][i]) {Lowcost[i]=lowcost[minp]+Cost[minp][i]; Minl[i]=min (minl[minp],l[i]); Maxl[i]=Max (maxl[minp],l[i]); } }}intMAP1[MAXN][MAXN];intANS[MAXN];intMain () {//freopen ("In.txt", "R", stdin); //freopen ("OUT.txt", "w", stdout);scanf ("%d%d",m,

POJ 1062 expensive offer shortest Circuit DescriptionThe young explorer came to an Indian tribe. There he fell in love with the daughter of the chief, so he went to ask for advice from the Chief. The chief promised to marry his daughter only when he used 10000 gold coins as a dowry. The explorer asked the chief to lower his requirements if he could not get so many gold coins. The chief said, "Well, if you can get me the high priest's leather jacket, I

I feel that I have not written the DFS Algorithm for a long time, because this topic retains and computes some values during the DFS process, so the program looks a little ugly. I will not talk much about my ideas. The current level is limited. The main reason for posting a blog is to write it to myself, as a proof of study for a period of time. Question address: http://poj.org/problem? Id = 1062 This is also a rare Chinese Question on poj, so it can

Code:#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. Pat (A) 1062. Talent and Virtue (struct sort)

Title Link: http://www.lightoj.com/volume_showproblem.php?problem=1062Test instructions: Two sticks obliquely on the wall, giving you the length of the stick and the height of their intersection distance from the ground, find the distance between the two wallsIdea: Direct enumeration distance of two points can beAC Code:#include #include #include #include #include #include #include #include #include #define MAXN 1010000#define LL Long Long#define LL __int64#define INF 0xFFFFFFF#define MEM (x) me

POJ on a rare Chinese question ...Idea: To create a map with a source point of 0, then the value of Map[0][n] represents the value of item n, and Map[i][j] represents the value of item J after substituting J for I. We think that the chief's commitment is node ' 1 ', then what we need to do is to find the minimum value of Map[0][1 by a series of operations, then we can see that this is a shortest path problem. The title also stipulates that the high level will not agree with the low level of exch

About years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people ' s talent and VI Rtue. According to He theory, a man being outstanding in both talent and virtue must is a "sage (sage)"; Being less excellent but with one's virtue outweighs talent can be called a "nobleman (gentleman)"; Being good in neither are a "fool Man (Fool)"; Yet a fool man was better than a "small man (villain)" who prefers talent than virtue.Now given the grades of talent and virtue

of things, because they won't get a lower price. Explorers now need your help so that he can marry his sweetheart with the least amount of gold. And what he's going to tell you is that in this tribe, hierarchy is very strong. No direct contact will be made between two persons with a status gap exceeding a certain limit, including transactions. He is a foreigner, so he can not be subject to these restrictions. But if he trades with a lower-level person, the higher-status person will not trade wi

online algorithm offline algorithm What is now said to you listen you do not necessarily be able to understand quickly. ” "But ... Didn't you say I could write it? Little Ho was puzzled. "So let's take it slow, and teach you the simplest way first." "said Little hi. "The simplest way ... If the amount of data is not large, I can completely find out the ancestors of these two people, and then take their intersection, and then find one of the generational the lowest one. "Little Ho thinks s

Test instructions: Provide a family tree asking about the public ancestor closest to two people.Idea: See the Great God code to learn this idea. Use map to save the son to his father's mapping, query, two people together back to the root, the middle of the meeting is the answer to the node. Use one to backtrack to the root, record the node on the path to access, and then use the other to go back to the root, as long as the first encounter has been visited nodes, is the answer.1#include 2#include

Title Link: http://www.lightoj.com/volume_showproblem.php?problem=1062Test instructions: Two ladders against the wall, one length x one length is Y, and the distance from the intersection to the ground is C. Find the distance between the bottom of the two ladders.Idea: The distance from the bottom of the two points T, to calculate t ', according to T and T ' size relationship update upper bound.Set the width to mid, then you can obtainC/SQRT (y^2-mid^2) + c/sqrt (x^2-mid ^2) = 1Code:#include #in

Chinese questionsProblem-solving ideas: The most source point of the Chiefs, with a structure to record each point to the chief of the shortest, and the level range, when updating the new node, you can determine the level range to update the#include #include #include #include #define N#define M 10010#define INF 0x3f3f3f3fusing namespace STD;structPeople {intPrice, level, X;} P[n];structchange{intNo, Price;} C[M];structState {intDis, high, low;} S[n];intN, M, CNT;intDis[n][n];BOOLVis[n];voidInit