百度之星复赛Astar Round3

拍照

树状数组(SB了)。求出静止状态下，每个点能看到多少个向右开的船c1[i]，多少个向左开的船c2[i]。

max{c1[i] + c2[j], (满足i <= j)  }即为答案。从后往前枚举i即可。

注意要离散化，否则会Tle。

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

 const int maxn =2e4;
 //c1 向右开的船
 int c1[maxn<<], c2[maxn<<];
 int s1[maxn<<], s2[maxn<<];

 int lowbit(int x){ return x&-x;}
 void add(int x, int d, int* c){//c[x]++;
     while(x <= (maxn<<)){
         c[x] += d;
         x += lowbit(x);
     }
 }
 void Add(int l, int r, int* c){
     add(l, , c);
     add(r+, -, c);
 }
 int sum(int x, int* c){
     int ret = ;
     while(x > ){
         ret += c[x];
         x -= (x&-x);
     }
     return ret;
 }

 int n;
 int ope[maxn], tot;
 struct p{
     int l, r, d;
     p(){}
     p(int l, int r, int d):l(l), r(r), d(d){}
 };
 p pp[maxn];

 int main(){
     int t, ca = ;scanf("%d", &t);
     while(t--){
         scanf("%d", &n);
         memset(c1, , sizeof(c1));
         memset(c2, , sizeof(c2));
         int l, r, x, y, z, d;

         tot = ;
         for(int i = ; i < n; i++){
             scanf("%d%d%d%d", &x, &y, &z, &d);
             l = y-z+maxn, r = x+z+maxn;
             ope[tot++] = l, ope[tot++] = r;
             pp[i] = p(l, r, d);
         }

         sort(ope, ope+tot);
         //tot = unique(ope, ope+tot)-ope;
         for(int i = ; i < n; i++){
             int l = lower_bound(ope, ope+tot, pp[i].l)-ope+;
             int r = lower_bound(ope, ope+tot, pp[i].r)-ope+;
             int d = pp[i].d;
             if(l <= r)
                 Add(l, r, (d == ? c1 : c2));
         }

         for(int i = ; i < (maxn<<); i++)
             s1[i] = sum(i, c1);
         for(int i = ; i < (maxn<<); i++)
             s2[i] = sum(i, c2);

         int ans = ;
         for(int i = (maxn<<)-; i > ; i--){
             s2[i] = max(s2[i], s2[i+]);
             ans = max(ans, s1[i]+s2[i]);
         }
         printf("Case #%d:\n%d\n", ca++, ans);
     }
     return ;
 }

更新：SB了，要树状数组干什么用。。反正是要求出 每个点 能看到的船只数，直接做一遍前缀和累加一下就可以了。。。树状数组都省了。

代码如下：

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

 const int maxn =2e4;
 int c1[maxn<<], c2[maxn<<];

 void Add(int l, int r, int* c){
     c[l]++, c[r+]--;
 }

 int n;
 int ope[maxn], tot;
 struct p{
     int l, r, d;
     p(){}
     p(int l, int r, int d):l(l), r(r), d(d){}
 };
 p pp[maxn];

 int main(){
     int t, ca = ;scanf("%d", &t);
     while(t--){
         scanf("%d", &n);
         memset(c1, , sizeof(c1));
         memset(c2, , sizeof(c2));
         int l, r, x, y, z, d;

         tot = ;
         for(int i = ; i < n; i++){
             scanf("%d%d%d%d", &x, &y, &z, &d);
             l = y-z+maxn, r = x+z+maxn;
             ope[tot++] = l, ope[tot++] = r;
             pp[i] = p(l, r, d);
         }

         sort(ope, ope+tot);
         //tot = unique(ope, ope+tot)-ope;
         for(int i = ; i < n; i++){
             int l = lower_bound(ope, ope+tot, pp[i].l)-ope+;
             int r = lower_bound(ope, ope+tot, pp[i].r)-ope+;
             int d = pp[i].d;
             if(l <= r)
                 Add(l, r, (d == ? c1 : c2));
         }

         for(int i = ; i < (maxn<<); i++)
             c1[i] += c1[i-], c2[i] += c2[i-];

         int ans = ;
         for(int i = (maxn<<)-; i > ; i--){
             c2[i] = max(c2[i], c2[i+]);
             ans = max(ans, c1[i]+c2[i]);
         }
         printf("Case #%d:\n%d\n", ca++, ans);
     }
     return ;
 }