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线段树 区间合并 单点修改 区间查询。又是1秒钟构思,差错查了好久... ... 发现一个int型的定义成了char型,打脸。
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <stack> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int maxn= 50000+10; int N,Q; char op[5]; int num; stack<int>S; bool flag[maxn]; struct SegTree { int lsum,rsum,msum; }segTree[4*maxn]; int fail,ans1,ans2; void pushUp(int rt,int len) { if(segTree[2*rt].lsum==len-len/2) segTree[rt].lsum=segTree[2*rt].lsum+segTree[2*rt+1].lsum; else segTree[rt].lsum=segTree[2*rt].lsum; if(segTree[2*rt+1].rsum==len/2) segTree[rt].rsum=segTree[2*rt].rsum+segTree[2*rt+1].rsum; else segTree[rt].rsum=segTree[2*rt+1].rsum; segTree[rt].msum=max(segTree[2*rt].msum,segTree[2*rt+1].msum); segTree[rt].msum=max(segTree[rt].msum,segTree[2*rt+1].lsum+segTree[2*rt].rsum); } void build(int l,int r,int rt) { if(l==r) { segTree[rt].lsum=segTree[rt].rsum=segTree[rt].msum=1; return; } int m=(l+r)/2; build(l,m,2*rt); build(m+1,r,2*rt+1); pushUp(rt,r-l+1); } void update(int info,int node,int l,int r,int rt) { if(l==r&&node==l) { segTree[rt].lsum=segTree[rt].rsum=segTree[rt].msum=info; return; } int m=(l+r)/2; if(node<=m) update(info,node,l,m,2*rt); else update(info,node,m+1,r,2*rt+1); pushUp(rt,r-l+1); } void quary1(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R) { ans1=ans1+segTree[rt].rsum; if(segTree[rt].rsum!=r-l+1) fail=1; return; } int m=(l+r)/2; if(R>m) quary1(L,R,m+1,r,2*rt+1); if(fail) return; if(L<=m) quary1(L,R,l,m,2*rt); if(fail) return; } void quary2(int L,int R,int l,int r,int rt) { if(L<=l&&r<=R) { ans2=ans2+segTree[rt].lsum; if(segTree[rt].lsum!=r-l+1) fail=1; return; } int m=(l+r)/2; if(L<=m) quary2(L,R,l,m,2*rt); if(fail) return; if(R>m) quary2(L,R,m+1,r,2*rt+1); if(fail) return; } int main() { // freopen("in.txt","r",stdin); while(~scanf("%d%d",&N,&Q)) { while(!S.empty()) S.pop(); build(1,N,1); for(int i=0;i<=N;i++) flag[i]=1; for(int i=1;i<=Q;i++) { scanf("%s",op); if(op[0]==‘D‘) { scanf("%d",&num); S.push(num); flag[num]=0; update(0,num,1,N,1); } else if(op[0]==‘Q‘) { scanf("%d",&num); if(!flag[num]) printf("%d\n",0); else { ans1=0,ans2=0; fail=0; if(num-1>=1) quary1(1,num-1,1,N,1); fail=0; if(num+1<=N) quary2(num+1,N,1,N,1); int ans=ans1+ans2+1; printf("%d\n",ans); } } else if(op[0]==‘R‘) { if(S.empty()) continue; num=S.top(); S.pop(); flag[num]=1; update(1,num,1,N,1); } } } return 0; }
HDU 1540 POJ 2892 Tunnel Warfare
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原文地址:http://www.cnblogs.com/zufezzt/p/5020997.html