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1635: [Usaco2007 jan]tallest Cow Highest cow time limit:5 Sec Memory limit:64 MBsubmit:383 solved:211[Submit] [Status] DescriptionFJ ' s n (1 There are N (1 Input* Line 1:four space-separated integers:n, I, H and R* Lines 2..r+1:two distinct space-separated integers A and B (1 Output* Lines 1..n:line I contains the maximum possible height of cow I.Sample Input9 3 5 51 35 34 33 79 8INPUT DETAILS:There is 9 cows, and the 3rd was the tallest wit
3396: [Usaco2009 jan]total Flow Current time limit:1 Sec Memory limit:128 MB submit:179 solved:73 [Submit] [Status] Description InputThe 1th line enters N, and then each row of n lines describes a water pipe, and the first two letters indicate the two ends of the water pipe (the uppercase and lowercase letters are not the same), and the latter integer represents the flow of the water pipe, and the flow does not exceed 1000.OutputAn integer t
1677: [Usaco2005 jan]sumsets summation time limit:5 Sec Memory limit:64 MBsubmit:626 solved:348[Submit] [Status] DescriptionFarmer John commanded his cows to search for different sets of numbers the sum to a given number. The cows use is numbers that is an integer power of 2. Here is the possible sets of numbers that sum to 7:1) 1+1+1+1+1+1+1 2) 1+1+1+1+1+2 3) 1+1+1+2+2 4) 1+1+1+4 5) 1+2+2+2 6) 1+2+4 Help FJ count all possible representations
1596: [Usaco2008 Jan] Telephone network time limit:10 Sec Memory limit:162 MBsubmit:601 solved:265[Submit] [Status] [Discuss] DescriptionFarmer John has decided to equip all his cows with mobile phones to encourage them to communicate with each other. For this reason, however, FJ had to select a radio tower in the N (1 Input* Line 1th: 1 integers, N* 2nd. N rows: 2 integers separated by spaces A, B, for two adjacent grassland numbersOutput* L
1635: [usaco Jan] maximum cow time limit: 5 sec memory limit: 64 MB Submit: 346 solved: 184 [Submit] [Status] Description FJ's n (1 There are N (1 Input * Line 1: four space-separated integers: N, I, H and R * Lines 2. R + 1: two distinct space-separated integers A and B (1 Output * Lines 1. N: line I contains the maximum possible height of cow I.Sample input9 3 5 5 1 3 5 3 4 3 3 7 9 8 Input details: There are 9 cows, and the 3rd is the tallest
3396: [usaco 2009 Jan] total flow time limit: 1 sec memory limit: 128 MB Submit: 45 solved: 27 [Submit] [Status] Description Input Enter N in the second line, and then n lines describe a pipe in each line. The first two English letters indicate the two ends of the pipe (the upper and lower case letters are different), and the last integer indicates the traffic of the pipe, the traffic will not exceed 1000. Output An integer representing the total tr
Label: SP strong file Div on 2014 problem BS ad Usaco 2014 Jan I. Question Overview Chinese question name Team balance Ski video Ski Resort Construction English question name Bteams Recording Skicourse Executable File Name Bteams Recording Skicourse Input File Name Bteams. In Recording. In Skicourse. In Output file name Bteams. Out R
Tags: des style blog HTTP color Io OS AR 1596: [usaco 162 Jan] Telephone Network Time Limit: 10 sec memory limit: MB Submit: 513 solved: 232 [Submit] [Status] Description Farmer John decided to equip all his cows with cell phones to encourage them to communicate with each other. However, for this reason, FJ must select some radio communication towers in the N (1 Input * Row 1st: 1 integer, n * Row 2nd. N: two integers, A and B, separated by spac
At first, the face was crazy.Later I thought of maintaining a left and right hand two pointers L and R. Represents the different kinds of digital It is clear that the leftmost, legal L increases with R and does not decrease.By the way discretization, remember the number of different kinds of numbers to calculate the answer.Time complexity O (n)1#include 2#include 3#include 4#include 5 using namespacestd;6 Const intmaxn=1e5+233;7 structzs{intV,id;} A[MAXN];8 intMP[MAXN],SM[MAXN];9 inti,j,n,m,ans,
For an edge x->y, if x cannot reach y after it is removed, it is required.First, the topological order is sorted out and then the first key is followed by the topological order of the end point, and the starting topological order is the second keyword from small to large plus edge.For each point, maintain a bitset that represents the point at which you are currently able to reach yourself.Time Complexity $o (\frac{nm}{32}) $.#include BZOJ3355: [Usaco2004 J
：if(i==1||lsh[i].x!=lsh[i-1].x)m+=2;Code:#include #include #include #include #define N 50500#define LS (note#define RS (noteusing namespace STD;structlsh{Long LongL,r,x;BOOL operatorConstLSH a)Const{returnxLong Long_l=0,Long Long_r=0,Long Long_x=0): L (_l), R (_r), X (_x) {}}lsh[n*2],q[n];intn,m,cnt;structsegment_tree{intL,r,c;} t[i*6*4];voidPushup (intNote) {s[note].c=s[note].c| ( S[LS].CAMP;S[RS].C);}voidBuildintNoteintLintR) {s[note].l=l,s[note].r=r;if(L==R)return;intMid=l+r>>1
intU = Q.front (), flag =0; - Q.pop (); - for(inti = Head[u]; I i =Nex[i]) - { - intU =Point[i]; - if(A[u]) in { -Flag =1; to Break; + } - } the if(flag)Continue; *cnt++; $ for(inti = head[u];i;i=Nex[i])Panax Notoginseng { - intK =Point[i]; the if(Visit[k])Continue; +VISIT[K] =1; A Q.push (k); the } + } - returnCNT; $ } $ intMain () -
;//Can Else MapA [b]=2;//No} for(i=1; i1001; i++) {inttemp=0; for(j=1; j1001; j + +) {if(MapI [j]==1) temp++; f[i][j]=f[i-1][j]+temp; } } for(i=1; i1001; i++) S[i].h=inf;intL,r; for(i=1; i1001; i++) { for(j=1; j1001; j + +) {if(MapI [j]==2) S[j]. H=0, S[j].h=inf;Else{S[j]. h++;if(MapI [j]==1) s[j].h=1;Elses[j].h++; } s[j].l=s[j].r=j; } l=1, r=0; for(j=1; j1001; j + +) { while(L while(S[j].l1, r=0; for(j=1001; j;j--) { while(L while(S[j].r>js[s[j].r].h>s[j]. H) s[j].r-
,intXinty) -{HP C; c.delta=-1; c.size=a.size+b.size; in if(! A.W) c.w=b.w,c.li=0, c.ri=b.ri,c.p=B.P; - Else if(! B.W) c.w=a.w,c.li=a.li,c.ri=0, c.p=A.P; to Else{c.li=a.li+ (a.li==a.size? B.li:0); +c.ri=b.ri+ (b.ri==b.size? A.ri:0); - if(A.WGT;=B.W) c.w=a.w,c.p=A.P; the Elsec.w=b.w,c.p=B.P; * if(a.ri+b.li>c.w| | (a.ri+b.li==c.wmid-a.ri+1C.P)) $c.w=a.ri+b.li,c.p=mid-a.ri+1;Panax Notoginseng}returnC; - } the voidInsertintIintXintYintSintTintk) +
not be used Dijkstra, because it has negative edge rights. But we carefully observed that the negative edge can only be a route, and the route will only connect two can not directly to the Unicom block, that is to say, we can run in the Unicom block DijkstraWhat about the link between block and block? This problem after the point is a DAG, then we can topology, with the topological processing blocks and blocks of the relationship between the block directly Dijkstra, then this problem is doneThe
,v,w; toscanf"%d%d%d",u,v,W); + Addedge (u,v,w); - } the } * $ intCheckintx)Panax Notoginseng { -queueint>que; thememset (Inque,0,sizeof(Inque)); +memset (DIS,0x7f,sizeof(DIS)); Adis[1]=0, inque[1]=1; theQue.push (1); + while(!que.empty ()) - { $ intHead=Que.front (); Que.pop (); $inque[head]=0; - for(intI=0; I) - { the intnowlen,to=e[head][i].to; - if(e[head][i].dis>x) nowlen=1;Elsenowlen=0;Wuyi if(dis[head]+nowlenDis[to
0x7fffffffusing namespacestd;structdata{intTo,next,c;} e[20001];intn,p,k,cnt,head[1001],dis[1001],ans=-1;BOOLvis[1001];voidInsertintUintVintW) {CNT++;e[cnt].to=v;e[cnt].c=w;e[cnt].next=head[u];head[u]=CNT;}BOOLSPFA (intx) { intsum,u,v; Queueint>Q; Q.push (1); memset (DIS,127/3,sizeof(DIS)); dis[1]=0; while(!Q.empty ()) {u=q.front (); Q.pop (); vis[u]=0; for(intI=head[u];i;i=E[i].next) {v=e[i].to; if(e[i].c>x) sum=dis[u]+1; Elsesum=Dis[u]; if(dis[v]>sum) {Dis[v]=sum; if(!Vis[v]) {Vis[v]=1;
is to ensure that the bottom-right vertex of each operation is 1, and that the matrix contains the highest number of 1.Then the strategy is already there. The key code is as follows:Read the data in advance and ask for prefixes andfor (;;){BOOL Flag=false;for (int i=1;i{for (int j=1;jif (a[i][j]== ' 1 ') {flag=true;break;}if (flag) break;}if (!flag) break;//The judgment matrix is all 0ans++;int maxx=0,x=0,y=0;for (int i=1;ifor (int j=1;jif (sum[i][j]>maxxa[i][j]== ' 1 ')//Find the lower right p
the successful round danceI have not read the question ... All kinds of spit groove 233, and then turned over the problem, want to see if there is no topic, the results of an open is "bare tarjin", "strong connectivity components", such as concise and crisp, alas, so, I also write to read it (actually did not write before). Oh, incredibly a, or 1a,2b young people happy more ...#include #includeusing namespacestd;structna{intX,y,ne; Na () {ne=0; }};intn,m,l[10001],r[10001],x,y,num=0, ans=0, top=
O (3^9) enumeration symbol:Success at the bottom qaq ... Fortunately, it didn't break the 100% ac rate.1#include 2#include 3#include 4 using namespacestd;5 intl[ One],r[ One];6 intfh[ One],num[ One],a[ One];7 intI,j,k,n,m,ans;8 9 intRaCharRx;TenInlineintRead () { OneRx=getchar (), ra=0; A while(rx'0'|| Rx>'9') rx=GetChar (); - while(rx>='0'rx'9') ra*=Ten, ra+=rx- -, Rx=getchar ();returnRA; - } theInlineintCalc () { - inti,j,k; - for(i=1; ii; - for(i=1; i){ + for
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