Codeforces Round #154 (Div. 2)

转载请注明出处，谢谢http://blog.csdn.net/ACM_cxlove?viewmode=contents    by---cxlove 

唔，又玩脱了，哎，太弱

A:水题

B：枚举+二分  nlgn

      维护两个指针 O(n)

C：范围不大，直接BFS可搞

      显然中间段不会左右移动，所以枚举某个中间行r，顺序为r1->r->r2，到了r2行才左右移动

       ps:卡了半天，从a->b与b->a是一样的，我为了简便，交换了下，sad

D：DP预处理出a的个数，dp[i][j]表示以i,j为右下角的矩阵里有多少个a

然后枚举两列，类似B题，维护两个指针

E：想歪了，显然x的优化级越高，完成的越早

   单调性啊，直接二分优化级，然后模拟一遍，记录x的完成时间

#include<iostream>  #include<cstdio>  #include<map>  #include<cstring>  #include<cmath>  #include<vector>  #include<algorithm>  #include<set>  #include<string>  #include<queue>  #define inf 1600005  #define M 40  #define N 200000 #define maxn 300005  #define eps 1e-12#define zero(a) fabs(a)<eps  #define Min(a,b) ((a)<(b)?(a):(b))  #define Max(a,b) ((a)>(b)?(a):(b))  #define pb(a) push_back(a)  #define mp(a,b) make_pair(a,b)  #define mem(a,b) memset(a,b,sizeof(a))  #define LL unsigned long long  #define MOD 1000000007#define lson step<<1#define rson step<<1|1#define sqr(a) ((a)*(a))  #define Key_value ch[ch[root][1]][0]  #define test puts("OK");  #define pi acos(-1.0)#define lowbit(x) ((-(x))&(x))#pragma comment(linker, "/STACK:1024000000,1024000000")  using namespace std;struct Node{    int s,t,p,id;    Node(){}    Node(int _t,int _s,int _p,int _i):t(_t),s(_s),p(_p),id(_i){}    bool operator<(const Node n)const{        return t<n.t;    }};struct work{    int s,p,id;    work(){}    work(int _s,int _p,int _i):s(_s),p(_p),id(_i){}    bool operator<(const work w)const{        return p<w.p;    }};vector<Node>a;int n,pos;LL T;LL ans[50005];LL check(int p){    a[pos].p=p;    vector<Node> b(a);    sort(b.begin(), b.end());    priority_queue<work>que;    while(!que.empty()) que.pop();    que.push(work(b[0].s,b[0].p,b[0].id));    LL st=0;    for(int i=1;i<n;i++){        st=b[i-1].t;        LL tmp=b[i].t-b[i-1].t;        while(!que.empty()&&tmp){            work u=que.top();            que.pop();            LL cnt=min(tmp,(LL)u.s);            st+=cnt;            u.s-=cnt;            tmp-=cnt;            if(u.s==0) ans[u.id]=st;            else que.push(u);        }        st=b[i].t;        que.push(work(b[i].s,b[i].p,b[i].id));    }    while(!que.empty()){        work u=que.top();        que.pop();        st+=u.s;        ans[u.id]=st;    }    return ans[pos];}int main(){    freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    while(scanf("%d",&n)!=EOF){        a.clear();        for(int i=0;i<n;i++){            int t,s,p;            scanf("%d%d%d",&t,&s,&p);            a.pb(Node(t,s,p,i));            if(a[i].p==-1) pos=i;        }        scanf("%I64d",&T);        int low=1,high=1e9,mid;        while(low<=high){            mid=(low+high)>>1;            LL tmp=check(mid);            if(tmp==T) break;            if(tmp<T) high=mid-1;            else low=mid+1;        }        printf("%d\n",mid);        for(int i=0;i<n-1;i++) printf("%I64d ",ans[i]);        printf("%I64d\n",ans[n-1]);    }    return 0;}