CF371 D Searching Rectangles

来源：互联网发布：淘宝卖家能给差评吗编辑：程序博客网时间：2024/05/23 15:41

基本思路就是二分　每条边分别二分求
c++11 用fflush(sdtout) 不行　囧啊

#include<bits/stdc++.h>using namespace std;int ans[2][4];int inp;int check(int x1,int y1,int x2,int y2) {    printf("? %d %d %d %d\n",x1,y1,x2,y2);    fflush(stdout);    scanf("%d",&inp);    return inp;}int has(int x1,int y1,int x2,int y2) {    if(x1 <= ans[0][0] && y1 <= ans[0][1] && x2 >= ans[0][2] && y2 >= ans[0][3]) return 1;    return 0;}void solve(int x1,int y1,int x2,int y2,int flag){    int l=x1, r=x2;      while(l<=r){        int mid = (l+r)>>1;        int num = check(mid,y1,x2,y2);        if(flag &&  has(mid,y1,x2,y2) ) num--;        if(num==1 || num==2)            ans[flag][0] = mid, l = mid+1;        else            r = mid-1;    }    x1 = ans[flag][0];//  printf("x1:%d\n",x1);    l=x1,r=x2;    while(l<=r){        int mid = (l+r)>>1;        int num = check(x1,y1,mid,y2);        if( flag && has(x1,y1,mid,y2) ) num--;        if(num==1 || num==2)            ans[flag][2]=mid, r = mid-1;        else            l = mid+1;    }    x2 = ans[flag][2];//  printf("x2;%d\n",x2);    l=y1,r=y2;    while(l<=r){        int mid = (l+r)>>1;        int num = check(x1,mid,x2,y2);        if(flag &&  has(x1,mid,x2,y2) ) num--;        if(num==1 || num==2)            ans[flag][1]=mid, l = mid+1;        else            r=mid-1;    }    y1 = ans[flag][1];//  printf("y1:%d\n",y1);    l=y1,r=y2;    while(l<=r){        int mid = (l+r)>>1;        int num = check(x1,y1,x2,mid);        if(flag &&  has(x1,y1,x2,mid) ) num--;        if(num==1 || num==2)            ans[flag][3]=mid, r=mid-1;        else            l=mid+1;    }    y2 = ans[flag][3];//  printf("y2:%d\n",y2);}int main(){    int n;    while(~scanf("%d",&n)) {        solve(1,1,n,n,0);        solve(1,1,n,n,1);        printf("!");        for(int i = 0; i < 2; ++i) {            for(int j = 0; j < 4; ++j) {                printf(" %d",ans[i][j]);            }         }printf("\n"); fflush(stdout);    }    return 0;}

0 0