Description
Now and then you play the following game with your friend. Your friend writes down a sequence consisting of zeroes and ones. You choose a continuous subsequence (for example the subsequence from the third to the fifth digit inclusively) and ask him, whether this subsequence contains even or odd number of ones. Your friend answers your question and you can ask him about another subsequence and so on. Your task is to guess the entire sequence of numbers. You suspect some of your friend's answers may not be correct and you want to convict him of falsehood. Thus you have decided to write a PRogram to help you in this matter. The program will receive a series of your questions together with the answers you have received from your friend. The aim of this program is to find the first answer which is provably wrong, i.e. that there exists a sequence satisfying answers to all the previous questions, but no such sequence satisfies this answer.Input
The first line of input contains one number, which is the length of the sequence of zeroes and ones. This length is less or equal to 1000000000. In the second line, there is one positive integer which is the number of questions asked and answers to them. The number of questions and answers is less or equal to 5000. The remaining lines specify questions and answers. Each line contains one question and the answer to this question: two integers (the position of the first and last digit in the chosen subsequence) and one Word which is either `even' or `odd' (the answer, i.e. the parity of the number of ones in the chosen subsequence, where `even' means an even number of ones and `odd' means an odd number).Output
There is only one line in output containing one integer X. Number X says that there exists a sequence of zeroes and ones satisfying first X parity conditions, but there exists none satisfying X+1 conditions. If there exists a sequence of zeroes and ones satisfying all the given conditions, then number X should be the number of all the questions asked.Sample Input
1051 2 even3 4 odd5 6 even1 6 even7 10 oddSample Output
3ps：和hdu 3038類似，根據(jù)d(i,j]+d(j,k]=d(i,k]，就可以把所有的點通過與根節(jié)點的關(guān)系聯(lián)系起來了但是這道題最大的問題不在這里，而是它的數(shù)據(jù)太大了，所以需要離散化一下，但是，從沒有用過hash離散的我到哪離散去0.0搜了一下離散化的博客，但是要么只是告訴離散的方法，要么就是離散化的代碼寫了幾百行0.0，就不能來個簡單點的最后還是看了本題的題解才略微明白了一點什么叫離散化可以看到本題中的區(qū)間范圍很大，但是輸入的點卻最多只有5000*2個，所以可以從這里入手，建立映射關(guān)系比如建立一個f(x)的映射關(guān)系，把每一個輸出的點都存入f(x)函數(shù)中，所以f(x)函數(shù)最多只有5000*2個數(shù)，而后我們查詢的時候只需要借用他們映射的位置，通過其位置來實現(xiàn)對數(shù)據(jù)的處理，也就是說我們所用的只是這5000*2個位置與數(shù)據(jù)的關(guān)系，這樣就實現(xiàn)了離散化（暫時我的離散化就是這樣的）代碼：
#include<stdio.h>#include<stdlib.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 10000+10struct node{ int u,v,w;} a[maxn];int pre[maxn],rank[maxn];int f[maxn];void init(int n){ for(int i=0; i<n; i++) { pre[i]=i; rank[i]=0; }}int Find(int x){ int temp=pre[x]; if(x==pre[x]) return x; pre[x]=Find(temp); rank[x]=(rank[x]+rank[temp])%2; return pre[x];}int main(){ int n,m; while(~scanf("%d%d",&n,&m)) { int i,cnt=0; char s[10]; for(i=1; i<=m; i++) { scanf("%d%d %s",&a[i].u,&a[i].v,s); a[i].u--; if(!strcmp(s,"even")) a[i].w=0; else a[i].w=1; f[cnt++]=a[i].u; f[cnt++]=a[i].v; } sort(f,f+cnt); n=unique(f,f+cnt)-f; init(n); int sum=0; for(i=1; i<=m; i++) { int x=lower_bound(f,f+n,a[i].u)-f;//查詢其位置 int y=lower_bound(f,f+n,a[i].v)-f;// int fx=Find(x),fy=Find(y); if(fx!=fy) { pre[fx]=fy; rank[fx]=(rank[y]+a[i].w-rank[x])%2;//通過其位置來建立關(guān)系 } else { if((abs(rank[x]-rank[y])%2)!=a[i].w) break; } sum++; } printf("%d/n",sum); } return 0;}
新聞熱點
疑難解答
