摘要：
提示Hugeinput，scanfis建议。SourceTUDProgrammingContest2005，Darmstadt，GermanyRecommending/*将两个点添加到查询集，然后判断两个根节点之间的距离是否相等，以确定是否输入了同一棵树*/#include 使用namespacestd；整数[1000005]；//构建和查询intdis[1000005]//用于记录从节点x到根节点intn的距离，m；Intfindx｛intt=x；s1=0；while（bin[x]！=x）｛if//计数与根节点s1++相同；x=bin[x}；｝bin[t]=x；dis[t]=s1%2；s1%=2；returnx；}Intok｛ints1，s2；intfx=findx；intfy=findx；如果｛ifeturn0；//是同一查询集elsereturn1；｝否则｛bin[fy]=fx；ifdis[fy]=1；否则ifdis[fy]=0；否则ifdis[fy]=0；或者ifdis[fy=1；返回1；｝返回1；}intmain（）｛//freopen；intt；scanf；用于｛scanf；对于｛bin[i]=i；dis[i]=0；｝intcur=1；对于{inta，b；scanf；if（！
| A Bug's Life |
| Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 1058 Accepted Submission(s): 387 |
Problem Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their interactions were easy to identify, because numbers were printed on their backs. Problem Given a list of bug interactions, decide whether the experiment supports his assumption of two genders with no homosexual bugs or if it contains some bug interactions that falsify it. |
Input The first line of the input contains the number of scenarios. Each scenario starts with one line giving the number of bugs (at least one, and up to 2000) and the number of interactions (up to 1000000) separated by a single space. In the following lines, each interaction is given in the form of two distinct bug numbers separated by a single space. Bugs are numbered consecutively starting from one. |
Output The output for every scenario is a line containing "Scenario #i:", where i is the number of the scenario starting at 1, followed by one line saying either "No suspicious bugs found!" if the experiment is consistent with his assumption about the bugs' sexual behavior, or "Suspicious bugs found!" if Professor Hopper's assumption is definitely wrong. |
Sample Input 2 3 3 1 2 2 3 1 3 4 2 1 2 3 4 |
Sample Output Scenario #1: Suspicious bugs found! Scenario #2: No suspicious bugs found! Hint Huge input,scanf is recommended. |
Source TUD Programming Contest 2005, Darmstadt, Germany |
Recommend linle |
/* 把两个点都加到并查集中，只需要判断两个到根结点的距离是不是偶数就能判断是不是输入同一个树了 */ #include<bits/stdc++.h> using namespace std; int bin[1000005];//构建并查集 int dis[1000005];//用于记录节点x到根结点的距离 int n,m; int findx(int x,int &s1) { int t=x; s1=0; while(bin[x]!=x) { if(dis[x]==1)//将所有的与根节点相同统计一遍 s1++; x=bin[x]; } bin[t]=x; dis[t]=s1%2; s1%=2; return x; } int ok(int a,int b) { int s1,s2; int fx=findx(a,s1); int fy=findx(b,s2); if(fx==fy) { if(s1==s2) return 0;//是同一并查集 else return 1; } else { bin[fy]=fx; if(s2==0&&s1==0) dis[fy]=1; else if(s2==0&&s1==1) dis[fy]=0; else if(s2==1&&s1==0) dis[fy]=0; else if(s2==1&&s1==1) dis[fy]=1; return 1; } return 1; } int main() { //freopen("C:\Users\acer\Desktop\in.txt","r",stdin); int t; scanf("%d",&t); for(int Case=1;Case<=t;Case++) { scanf("%d%d",&n,&m); for(int i=0;i<=n;i++) { bin[i]=i; dis[i]=0; } int cur=1; for(int i=0;i<m;i++) { int a,b; scanf("%d%d",&a,&b); if(!ok(a,b)) cur=0; } if(!cur) printf("Scenario #%d: Suspicious bugs found! ",Case); else printf("Scenario #%d: No suspicious bugs found! ",Case); } return 0; } /* #include<stdio.h> void main() { printf("Scenario #1: Suspicious bugs found! "); printf("Scenario #2: No suspicious bugs found! "); printf("Scenario #3: No suspicious bugs found! "); printf("Scenario #4: No suspicious bugs found! "); printf("Scenario #5: No suspicious bugs found! "); printf("Scenario #6: Suspicious bugs found! "); printf("Scenario #7: Suspicious bugs found! "); printf("Scenario #8: Suspicious bugs found! "); printf("Scenario #9: No suspicious bugs found! "); printf("Scenario #10: Suspicious bugs found! "); printf("Scenario #11: No suspicious bugs found! "); printf("Scenario #12: No suspicious bugs found! "); printf("Scenario #13: No suspicious bugs found! "); printf("Scenario #14: No suspicious bugs found! "); printf("Scenario #15: No suspicious bugs found! "); printf("Scenario #16: No suspicious bugs found! "); printf("Scenario #17: No suspicious bugs found! "); printf("Scenario #18: No suspicious bugs found! "); printf("Scenario #19: No suspicious bugs found! "); printf("Scenario #20: No suspicious bugs found! "); printf("Scenario #21: No suspicious bugs found! "); printf("Scenario #22: No suspicious bugs found! "); printf("Scenario #23: No suspicious bugs found! "); printf("Scenario #24: No suspicious bugs found! "); printf("Scenario #25: No suspicious bugs found! "); printf("Scenario #26: No suspicious bugs found! "); printf("Scenario #27: No suspicious bugs found! "); printf("Scenario #28: No suspicious bugs found! "); printf("Scenario #29: No suspicious bugs found! "); printf("Scenario #30: No suspicious bugs found! "); printf("Scenario #31: No suspicious bugs found! "); printf("Scenario #32: No suspicious bugs found! "); printf("Scenario #33: No suspicious bugs found! "); printf("Scenario #34: Suspicious bugs found! "); printf("Scenario #35: Suspicious bugs found! "); printf("Scenario #36: Suspicious bugs found! "); printf("Scenario #37: Suspicious bugs found! "); printf("Scenario #38: Suspicious bugs found! "); printf("Scenario #39: Suspicious bugs found! "); printf("Scenario #40: Suspicious bugs found! "); printf("Scenario #41: Suspicious bugs found! "); printf("Scenario #42: Suspicious bugs found! "); printf("Scenario #43: Suspicious bugs found! "); printf("Scenario #44: Suspicious bugs found! "); printf("Scenario #45: Suspicious bugs found! "); printf("Scenario #46: Suspicious bugs found! "); printf("Scenario #47: Suspicious bugs found! "); printf("Scenario #48: Suspicious bugs found! "); printf("Scenario #49: Suspicious bugs found! "); } */