Haybale Guessing
| Time Limit: 1000MS | Memory Limit: 65536K | |
Description
The cows, who always have an inferiority complex about their intelligence, have a new guessing game to sharpen their brains.
A designated 'Hay Cow' hides behind the barn and creates N (1 ≤ N ≤ 1,000,000) uniquely-sized stacks (conveniently numbered 1..N) of hay bales, each with 1..1,000,000,000 bales of hay.
The other cows then ask the Hay Cow a series of Q (1 ≤ Q ≤ 25,000) questions about the the stacks, all having the same form:
What is the smallest number of bales of any stack in the range of stack numbers Ql..Qh (1 ≤ Ql ≤ N; Ql ≤ Qh ≤ N)?
The Hay Cow answers each of these queries with a single integer A whose truthfulness is not guaranteed.
Help the other cows determine if the answers given by the Hay Cow are self-consistent or if certain answers contradict others.
Input
* Line 1: Two space-separated integers: N and Q
* Lines 2..Q+1: Each line contains three space-separated integers that represent a single query and its reply: Ql, Qh, and A
Output
* Line 1: Print the single integer 0 if there are no inconsistencies among the replies (i.e., if there exists a valid realization of the hay stacks that agrees with all Q queries). Otherwise, print the index from 1..Q of the earliest query whose answer is inconsistent with the answers to the queries before it.
Sample Input
20 4 1 10 7 5 19 7 3 12 8 11 15 12
Sample Output
3
分析：二分答案，然后按A从大到小遍历；
对于相同的值A，判断区间交是否成立，然后覆盖区间并；
覆盖的过程可以用线段树或并查集实现；
注意离散化过程排除[a,b]!=[a,a]U[b,b](a>b+1);
可以排序后再在相邻的差值>1的点之间插值；
代码：
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <bitset> #include <map> #include <queue> #include <stack> #include <vector> #include <cassert> #define rep(i,m,n) for(i=m;i<=n;i++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define sys system("pause") const int maxn=1e6+10; const int N=2e5+10; using namespace std; int id(int l,int r){return l+r|l!=r;} ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;} int n,m,k,t,mi[maxn<<1],tag[maxn<<1],q,ql[maxn],qr[maxn],qt[maxn],ip[maxn],cnt; double a[maxn]; bool cmp(int x,int y){return qt[x]>qt[y];} void pdw(int x,int y,int rt) { int mid=x+y>>1,ls=id(x,mid),rs=id(mid+1,y); mi[ls]=mi[rs]=tag[rt]; tag[ls]=tag[rs]=tag[rt]; tag[rt]=0; } void pup(int x,int y,int rt) { int mid=x+y>>1; mi[rt]=min(mi[id(x,mid)],mi[id(mid+1,y)]); } void init(int l,int r,int rt) { mi[rt]=0; tag[rt]=0; if(l==r)return; int mid=l+r>>1; init(l,mid,id(l,mid)); init(mid+1,r,id(mid+1,r)); } void upd(int x,int y,int z,int l,int r,int rt) { if(x==l&&y==r) { mi[rt]=z; tag[rt]=z; return; } int mid=l+r>>1; if(tag[rt])pdw(l,r,rt); if(y<=mid)upd(x,y,z,l,mid,id(l,mid)); else if(x>mid)upd(x,y,z,mid+1,r,id(mid+1,r)); else upd(x,mid,z,l,mid,id(l,mid)),upd(mid+1,y,z,mid+1,r,id(mid+1,r)); pup(l,r,rt); } int gao(int x,int y,int l,int r,int rt) { if(x==l&&y==r)return mi[rt]; int mid=l+r>>1; if(tag[rt])pdw(l,r,rt); if(y<=mid)return gao(x,y,l,mid,id(l,mid)); else if(x>mid)return gao(x,y,mid+1,r,id(mid+1,r)); else return min(gao(x,mid,l,mid,id(l,mid)),gao(mid+1,y,mid+1,r,id(mid+1,r))); } bool ok(int x) { int i,j; init(1,cnt,id(1,cnt)); rep(i,1,x)ip[i]=i; sort(ip+1,ip+x+1,cmp); for(i=1;i<=x;) { int l=ql[ip[i]],r=qr[ip[i]],xl=l,xr=r; j=i; while(j+1<=x&&qt[ip[j+1]]==qt[ip[i]])l=max(l,ql[ip[j+1]]),r=min(r,qr[ip[j+1]]),xl=min(xl,ql[ip[j+1]]),xr=max(xr,qr[ip[j+1]]),j++; if(l>r)return false; if(gao(l,r,1,cnt,id(1,cnt))>qt[ip[i]])return false; upd(xl,xr,qt[ip[i]],1,cnt,id(1,cnt)); i=j+1; } return true; } int main() { int i,j; scanf("%d%d",&n,&q); rep(i,1,q) { scanf("%d%d%d",&ql[i],&qr[i],&qt[i]); if(ql[i]>qr[i])swap(ql[i],qr[i]); a[++cnt]=ql[i],a[++cnt]=qr[i]; } sort(a+1,a+cnt+1); for(i=cnt;i>=1;i--)if(a[i]>a[i-1]+1)a[++cnt]=a[i-1]+1; sort(a+1,a+cnt+1); cnt=unique(a+1,a+cnt+1)-a-1; rep(i,1,q) { ql[i]=lower_bound(a+1,a+cnt+1,ql[i])-a; qr[i]=lower_bound(a+1,a+cnt+1,qr[i])-a; } int l=1,r=q,ret; while(l<=r) { int mid=l+r>>1; if(ok(mid))ret=mid,l=mid+1; else r=mid-1; } assert(ret>=1&&ret<=q); printf("%d\n",ret+1<=q?ret+1:0); return 0; }
