[UVALive7261]A

题目链接：https://vjudge.net/problem/UVALive-7261

题意略

三个步骤：

1.二分满足左边绿洲面积大于等于右边绿洲面积，并且使左边面积尽可能大的分割线位置。

2.判断这个分割线是否包含于任何一个绿洲中，如果包含，那么直接输出结果就行，否则：

3.从此坐标向右扫描，找到下一个绿洲的左边界，此为答案。

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef struct Sq { LL x, y, w, h; }Sq;
const int maxn = 10100;
LL n, k;
Sq sq[maxn];
LL sum;
map<LL, int> vis;

bool ok1(LL val) {
    LL a = 0;
    for(int i = 0; i < k; i++) {
        if(sq[i].x + sq[i].w <= val) a += sq[i].w * sq[i].h;
        else if(sq[i].x <= val && val <= sq[i].x+sq[i].w) {
            a += sq[i].h * (val - sq[i].x);
        }
    }
    LL b = sum - a;
    return a >= b;
}

int main() {
    // freopen("in", "r", stdin);
    int T;
    LL x, y, w, h;
    scanf("%d", &T);
    while(T--) {
        scanf("%lld%lld",&n,&k);
        sum = 0; vis.clear();
        vis[n] = 1;
        for(int i = 0; i < k; i++) {
            scanf("%lld%lld%lld%lld",&x,&y,&w,&h);
            sq[i] = Sq{x,y,w,h};
            vis[x] = 1;
            sum += w * h;
        }
        LL lo = 0, hi = n;
        LL ret = 0;
        while(lo <= hi) {
            LL mid = (lo + hi) >> 1;
            if(ok1(mid)) {
                ret = mid;
                hi = mid - 1;
            }
            else lo = mid + 1;
        }
        bool flag = 0;
        for(int i = 0; i < k; i++) {
            if(sq[i].x <= ret && ret < sq[i].x+sq[i].w) {
                flag = 1;
                break;
            }
        }
        if(flag) {
            printf("%lld
", ret);
            continue;
        }
        LL remain = 0;
        for(LL i = ret; i <= n; i++) {
            if(vis.find(i) != vis.end()) {
                remain = i - ret;
                break;
            }
        }
        printf("%lld
", ret+remain);
    }
    return 0;
}