Codeforces Round #310 (Div. 1) C. Case of Chocolate (线段树)

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题目地址：传送门
这题虽然是DIV1的C。。但是挺简单的。。只要用线段树分别维护一下横着和竖着的值就可以了，先离散化再维护。每次查找最大的最小值<=tmp的点，可以直接在线段树里搜，也可以二分去找。
代码如下：
#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>#include <map>#include <set>#include <stdio.h>#include <time.h>using namespace std;#define LL __int64#define pi acos(-1.0)//#pragma comment(linker, "/STACK:1024000000")#define root 0, cnt-1, 1#define lson l, mid, rt<<1#define rson mid+1, r, rt<<1|1const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;const int MAXN=400000+10;int a[MAXN], c[MAXN], cnt, ha[MAXN];int Min[2][MAXN<<2];struct node{        int x, y, f;}fei[MAXN];void PushUp(int f, int rt){        Min[f][rt]=min(Min[f][rt<<1],Min[f][rt<<1|1]);}void Update(int f, int p, int x, int l, int r, int rt){        if(l==r){                Min[f][rt]=x;                return ;        }        int mid=l+r>>1;        if(p<=mid) Update(f,p,x,lson);        else Update(f,p,x,rson);        PushUp(f,rt);}int seach(int f, int rr, int x, int l, int r, int rt){        if(l==r){                if(Min[f][rt]<=x) return l;                return -1;        }        int ans=-1, mid=l+r>>1;        if(rr>mid&&Min[f][rt<<1|1]<=x) ans=seach(f,rr,x,rson);        if(ans!=-1) return ans;        if(Min[f][rt<<1]<=x) ans=seach(f,rr,x,lson);        return ans;}int BS(int x){        int low=0, high=cnt-1, mid;        while(low<=high){                mid=low+high>>1;                if(c[mid]==x) return mid;                else if(c[mid]>x) high=mid-1;                else low=mid+1;        }}int main(){        int n, q, i, j, x, y, tmpx, tmpy, z;        char ch[3];        while(scanf("%d%d",&n,&q)!=EOF){                for(i=0;i<q;i++){                        scanf("%d%d",&fei[i].x,&fei[i].y);                        scanf("%s",ch);                        if(ch[0]=='U'){                                a[i<<1]=fei[i].x;                                a[i<<1|1]=fei[i].y;                                fei[i].f=0;                        }                        else{                                a[i<<1]=fei[i].x;                                a[i<<1|1]=fei[i].y;                                fei[i].f=1;                        }                }                memset(ha,0,sizeof(ha));                sort(a,a+2*q);                c[0]=a[0];                cnt=1;                for(i=1;i<2*q;i++){                        if(a[i]!=a[i-1]){                                c[cnt++]=a[i];                        }                }                memset(Min,INF,sizeof(Min));                for(i=0;i<q;i++){                        tmpx=BS(fei[i].x);                        tmpy=BS(fei[i].y);                        if(ha[tmpx]){                                puts("0");                                continue ;                        }                        ha[tmpx]=1;                        if(fei[i].f){                                if(tmpx==0){                                        printf("%d\n",fei[i].x);                                        Update(0,tmpy,0,root);                                        continue ;                                }                                z=seach(1,tmpx,tmpy,root);                                if(z==-1){                                        printf("%d\n",fei[i].x);                                        Update(0,tmpy,0,root);                                        continue ;                                }                                printf("%d\n",fei[i].x-c[z]);                                Update(0,tmpy,z+1,root);                        }                        else{                                if(tmpy==0){                                        printf("%d\n",fei[i].y);                                        Update(1,tmpx,0,root);                                        continue ;                                }                                z=seach(0,tmpy,tmpx,root);                                if(z==-1){                                        printf("%d\n",fei[i].y);                                        Update(1,tmpx,0,root);                                        continue ;                                }                                printf("%d\n",fei[i].y-c[z]);                                Update(1,tmpx,z+1,root);                        }                }        }        return 0;}
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