标签:show vector color out air names tac 问题 ace
题意:中文题
思路:大概就是尺取法,先按l从小到大再按r从大到小对区间排序,L,R表示当前计算的可行的区间,依次取每个区间,然后更新L,R,每次计算答案取最大值。。。数据有问题,wa了一天,不愧为百(垃)度(圾)之星
AC代码:
#include "iostream" #include "string.h" #include "stack" #include "queue" #include "string" #include "vector" #include "set" #include "map" #include "algorithm" #include "stdio.h" #include "math.h" #pragma comment(linker, "/STACK:102400000,102400000") #define ll long long #define endl ("\n") #define bug(x) cout<<x<<" "<<"UUUUU"<<endl; #define mem(a,x) memset(a,x,sizeof(a)) #define mp(x,y) make_pair(x,y) #define pb(x) push_back(x) #define ft first #define sd second #define lrt (rt<<1) #define rrt (rt<<1|1) using namespace std; const long long INF = 1e18+1LL; const int inf = 1e9+1e8; const int N=1e5+100; const ll mod=1e9+7; struct Node{ ll l, r; bool friend operator< (Node a, Node b){ if(a.l==b.l) return a.r>b.r; return a.l<b.l; } }a[N],x; queue<Node> Q; int main(){ ll n,m; while(scanf("%I64d %I64d",&n, &m)!=EOF){ for(int i=1; i<=n; ++i){ scanf("%I64d %I64d",&a[i].l, &a[i].r); } sort(a+1,a+1+n); while(!Q.empty()) Q.pop(); ll L=0, R=-1,ans=0; for(int i=1; i<=n; ++i){ if(a[i].l-1>R){ if(a[i].l-R-1<=m){ m-=a[i].l-R-1; x.l=R+1, x.r=a[i].l-1; Q.push(x); R=a[i].r; ans=max(ans,m+R-L+1);// cout<<m+R-L+1<<" 1"<<endl; } else{ while(!Q.empty()){ x=Q.front(); Q.pop(); L=x.r+1; ll c=x.r-x.l+1; m+=c; if(a[i].l-R-1<=m) break; } if(a[i].l-R-1<=m){ m-=a[i].l-R-1; x.l=R+1, x.r=a[i].l-1; Q.push(x); R=a[i].r; ans=max(ans,m+R-L+1); //cout<<m+R-L+1<<" 2"<<endl; } else{ x.l=a[i].l-m, x.r=a[i].l-1; Q.push(x); m=0; L=x.l, R=a[i].r; ans=max(ans, R-L+1); //cout<<m+R-L+1<<" 3"<<endl; } } } else{ //cout<<m+R-L+1<<" 4"<<endl; if(a[i].r>R){ R=a[i].r; ans=max(ans,m+R-L+1); } else ans=max(ans,m+R-L+1); } } cout<<ans<<endl; } return 0; }
标签:show vector color out air names tac 问题 ace
原文地址:http://www.cnblogs.com/max88888888/p/7354427.html