The main idea of the title: There is a 0,1 string of length n, and m pieces of information are given. Each piece of information indicates that the number of 1s in the middle of the xth to yth characters is even or odd. If the k+ 1 is the first time it contradicts the previous words, output k;
Idea: If the number of 1s between x and y is an even number, then the parity of 1 in 1~x and 1~y is the same; if the number of 1s between x and y is an odd number, then 1~x and 1~y have the same parity. The parity of 1 in 1~y is the same. Join the same set with the same parity
eg: The data range is large, discretize first
#include<stdio.h> #include<algorithm> #define N 100006 using namespace std; struct Node { int u,v; char str[10]; }that[N]; int a[N*2],n,q,cnt; int pre[N*2],r[N*2]; int Bin(int x) { int low=0,high=cnt-1,mid; while(low<=high) { mid=(low+high)/2; if(a[mid]==x) return mid; if(a[mid]<x) low=mid+1; else high=mid-1; } return 0; } int find(int x) { if(x!=pre[x]) { int f=pre[x]; pre[x]=find(pre[x]); r[x]=r[x]^r[f]; } return pre[x]; } intmain () { while(scanf("%d",&n)!=EOF) { scanf("%d",&q); cnt=0; for(int i=0;i<q;i++) { scanf("%d%d%s",&que[i].u,&que[i].v,&que[i].str); that[i].u -- ; a [cnt ++] = que [i] .v; a [cnt ++] = que [i] .u; } sort(a,a+cnt); cnt=unique(a,a+cnt)-a; for(int i=0;i<cnt;i++) pre[i]=i,r[i]=0; int ans=0; for(int i=0;i<q;i++) { int u=Bin(que[i].u),v=Bin(que[i].v); int ra=find(u),rb=find(v); if(ra==rb) { if(r[u]==r[v]&&que[i].str[0]=='o') break; if(r[u]!=r[v]&&que[i].str[0]=='e') break; years ++ ; } else { if(que[i].str[0]=='o') { for [ra] = rb; r [ra] = r [u] ^ r [v] ^ 1 ; } else { for [ra] = rb; r [ra] = r [u] ^ r [v]; } years ++ ; } } printf("%d\n",ans); } return 0; }