vostro 1015

) - { -vectorint>v; - for(intj=i+1; j) A if(mp[q[i]][q[j]]==1) V.push_back (Q[j]); + for(intj=1; J) the if(mp[v[0]][v[j]]==0)return false; - } $ return true; the } the the intMain () the { - //freopen ("In.txt", "R", stdin); in while(SCANF ("%d%d", n,m)! =EOF) the { the if(n==0 m==0) Break; Abouttot=0; thememset (head,-1,sizeof(head)); theMemset (MP,0,sizeof(MP)); the while(m--) + { - intu,v;

The reverse is thought of as all planets are destroyed and then restored to the planet.The number of connected blocks is used and checked for maintenance, each recovery cannot be counted as destroyed planet .#include #includeusing namespacestd;intn,m,k,fa[400001],x,y;vectorint> g[400001];intans[400001];BOOLd[400001];intv[400001];BOOLvis[400001];intGfintx) { if(fa[x]==x)returnx; returnfa[x]=GF (fa[x]);}intMain () {scanf ("%d%d",n,m); for(intI=0; ii; for(intI=1; i) {scanf ("%d%d",x,y);

The key point of the topic is that after reading it, we carry it backwards and check the set operation. Because the check set does not support the delete operation, it is read in. In order, it is possible to reverse the operation.#include #include#includeusing namespacestd; vectorint>edge[400005];intx,y,n,m;intk,ban[200005];inteax400005];BOOLvis[400005];inttot;intask[200005]; intFindintx) {if(X!=fa[x]) fa[x]=find (fa[x]); returnfa[x];} voidUnintXinty) {intFX =find (x); intFY =find (y); //printf

; i){ $scanf"%d%d",q[i].u,q[i].v); $ } - intK; -scanf"%d",k); the for(intI=1; i){ -scanf"%d",a[i]);Wuyivis[a[i]]=1; the } -tot=n-K; Wu for(intI=0; i){ - if(vis[q[i].u]| | VIS[Q[I].V]) {//if one side is removed, it will be two points to each other in the non-line array (two-way storage edge), otherwise the direct edge About UN[Q[I].U].PB (Q[I].V); $ UN[Q[I].V].PB (q[i].u); - } - Else { - Unite (Q[I].U,Q[I].V); A } + } theans[k+1]=tot; -

in all possible cases according to the dictionary order of the largest casesimple analysis can be seen, is a combinatorial problem, the maximum size of the problem solution is 12 to 5, is 12*11*10*9*8*7, and the minimum size is 5 take 5, so should use enumeration method can be done. However, before enumerating the first sequence, you can guarantee that the output is the largest one that meets the requirements .Violent water too! The simple dictionary order problem got the rookie blogger for mor

The title describes the sum of the following three numbers, preserving the 1~b of the sum of the squares and 1~c of the 2-bit decimal 1~a and the input a B c output 1+2+...+a + 1^2+2^2+...+b^2 + 1/1+1/2+...+1/c Sample input100 50 10Sample output47977.931 intMainintargcChar Const*argv[])2 {3 4 intA, B, C, I;5 floatS1 =0, S2 =0, S3 =0;6 7scanf"%d%d%d", a, b, c);8 9 for(i =1; I )Ten { OneS1 + =i; A } - - for(i =1; I ) the { -S2 + = i *i; - } - + for(i =1

And made a mistake of ZZ ...It is important to note that the destroyed planet is not a unicom block (probably only I do it =)Offline down time back, it becomes a map to add the planet, that is, with and check set maintenance Unicom, in the TOT variable record the current answer, each plus a planet on the tot++, each merge a unicom block on tot--Note that a planet that has never been destroyed should be added to the map before the clock is back.#include Bzoj

Test.cpp: Defines the entry point for a console application. #include "stdafx.h" #include #include #include #include # Include #include #include #include using namespace std; struct Stu {char num[10]; int de; int Cai; }; int Cmp1

=== Op Tech briefing, 2002/11/02 CST === "The item is locked in a Klein safe behind a painting in the second-floor library. klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in World War II. fortunately

Safecracker Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)Total submission (s): 3713 accepted submission (s): 1919 Problem description = Op Tech briefing, 2002/11/02 CST = "The item is locked in a Klein safe behind

In a tree-like country, two Princes want to set up their own capitals. The cities around the capital choose the capital near which they are loyal (that is, the prince gets the cities, when the distance is the same, select the prince's loyalty), the

A king will give his territory to two sons. The King's territory is a tree, N knots, and the N-1 side connects these knots, now, the son of large and small wants to select a node as his capital. In addition to the two sons, other cities (nodes) will

Problem description = Op Tech briefing, 2002/11/02 CST = "The item is locked in a Klein safe behind a painting in the second-floor library. klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in World War

Safe or unsafe time limit: 2000/1000 ms (Java/other) memory limit: 32768/32768 K (Java/other) total submission (s): 18 accepted submission (s ): 13 problem descriptionjavac ++ saw an interesting thing while reading books on the computer one day!

Edit distance time limit: 1 second memory limit: 64 m Problem description Given a string, an edit script is a set of instructions to turn itinto another string. There areThree kinds of operations in an edit script: Add: output one character.

The Song Dynasty historian Sima Guang in "the Capital Governance Tong Jian" has a famous "The Moral Only Theory": "is therefore only then the virtuous all says the sage, only then the German and the death is the Fool, the German wins only then is

Safecracker Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)Total submission (s): 8183 accepted submission (s): 4143Problem description = Op Tech briefing, 2002/11/02 CST = "The item is locked in a Klein safe behind a

URAL_1015 Cubes have only three essentially different rotations. Therefore, we can store the initial states of all dice and the derivative states after rotation in a hash table, each time you can use bfs or dfs to simulate rotating a dice, you can

# Include # Include # Include # Include Int CMP (const void * a, const void * B){Return * (char *) B-* (char *);}Int main (){Int target, Len, C [50000], flag, I, j, W, R, Q;Char A [50000];While (scanf ("% d % s", & target, )! = EOF){If (target = 0 &

To and fro # Include # include # include # include using namespace STD; int main () {string shuru; int column; // ifstream ifs ("shuju.txt"); While (CIN> column & column! = 0) {CIN> shuru; // ifs. ignore (); // Getline (IFS, shuru); char