題目鏈接：https://vjudge.net/problem/UVALive-7261

題意略

三個步驟：

1.二分滿足左邊綠洲面積大於等於右邊綠洲面積，並且使左邊面積盡可能大的分割線位置。

2.判斷這個分割線是否包含於任何一個綠洲中，如果包含，那麽直接輸出結果就行，否則：

3.從此坐標向右掃描，找到下一個綠洲的左邊界，此為答案。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 typedef struct Sq { LL x, y, w, h; }Sq;
 6 const
 
 int maxn = 10100;
 7 LL n, k;
 8 Sq sq[maxn];
 9 LL sum;
10 map<LL, int> vis;
11 
12 bool ok1(LL val) {
13     LL a = 0;
14     for(int i = 0; i < k; i++) {
15         if(sq[i].x + sq[i].w <= val) a += sq[i].w * sq[i].h;
16         else if(sq[i].x <= val && val <= sq[i].x+sq[i].w) {
 
17             a += sq[i].h * (val - sq[i].x);
18         }
19     }
20     LL b = sum - a;
21     return a >= b;
22 }
23 
24 int main() {
25     // freopen("in", "r", stdin);
26     int T;
27     LL x, y, w, h;
28     scanf("%d", &T);
29     while(T--) {
30         scanf("%lld%lld",&n,&k);
 
31         sum = 0; vis.clear();
32         vis[n] = 1;
33         for(int i = 0; i < k; i++) {
34             scanf("%lld%lld%lld%lld",&x,&y,&w,&h);
35             sq[i] = Sq{x,y,w,h};
36             vis[x] = 1;
37             sum += w * h;
38         }
39         LL lo = 0, hi = n;
40         LL ret = 0;
41         while(lo <= hi) {
42             LL mid = (lo + hi) >> 1;
43             if(ok1(mid)) {
44                 ret = mid;
45                 hi = mid - 1;
46             }
47             else lo = mid + 1;
48         }
49         bool flag = 0;
50         for(int i = 0; i < k; i++) {
51             if(sq[i].x <= ret && ret < sq[i].x+sq[i].w) {
52                 flag = 1;
53                 break;
54             }
55         }
56         if(flag) {
57             printf("%lld\n", ret);
58             continue;
59         }
60         LL remain = 0;
61         for(LL i = ret; i <= n; i++) {
62             if(vis.find(i) != vis.end()) {
63                 remain = i - ret;
64                 break;
65             }
66         }
67         printf("%lld\n", ret+remain);
68     }
69     return 0;
70 }

[UVALive7261]A - Xiongnu's Land （二分）