2019杭电多校第十场部分题解(1003,1005,1009)

1003: Valentine’s Day

贪心选择概率大的礼物，如果能使得1次惊喜的几率变大则加入，否则结束。（凭感觉莽的，没有仔细证明

#include
#define ll long long
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int maxn = 1e4 + 50;
double p[maxn];
int n;
int main()
{
	int T;cin>>T;
	while(T--){
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
             scanf("%lf", &p[i]);
        }
        sort(p, p+n);
        double p0 = 1.0, p1 = 0.0;
        for(int i = n-1; i >= 0; --i){
            if(p0-p1 > 0){
                p1 = p1 + (p0-p1)*p[i];
                p0 = p0*(1-p[i]);
            }
            else break;
        }
        printf("%.8f\n", p1);
	}
}

1005：Welcome Party

按x排序，枚举每一个人作为最大的x，设当前枚举到i，则i后面的人只能选y（因为i后面的人x都大于等于i的x），如果后面的最大的y大于等于当前x，那么差距无法再减少了。如果比当前x小，那么看i前面有没有能缩小差距的y。

#include
#define ll long long
using namespace std;
const int maxn = 1e5 + 50;
struct node{
    ll x, y;
    bool operator < (const node& a) const{return x < a.x;};
}e[maxn];
set<ll> s;
int n;
ll mx[maxn];
int main()
{
	int T;cin>>T;
	while(T--){
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%lld%lld", &e[i].x, &e[i].y);
        }
        sort(e, e+n);
        mx[n] = 0;
        for(int i = n-1; i >= 0; --i){
            mx[i] = max(e[i].y, mx[i+1]);
        }
        s.clear();
        ll ans = abs(e[0].x - mx[1]);
        s.insert(e[0].y);
        for(int i = 1; i < n; ++i){
            ll t;
            if(i < n-1 && e[i].x <= mx[i+1]){
                t = mx[i+1] - e[i].x;
            }
            else{
                if(i < n-1) t = e[i].x - mx[i+1];
                else t = 2e18;
                set<ll>::iterator it = s.lower_bound(e[i].x);
                if(it != s.end()){
                    t = min(t, *it - e[i].x);
                }
                if(it != s.begin()){
                    it--;
                    t = min(t, e[i].x - *it);
                }
            }
            ans = min(ans, t);
            s.insert(e[i].y);
        }
        printf("%lld\n", ans);
	}
}

1009：Block Breaker

按题意模拟

#include
#define ll long long
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int maxn = 2e3 + 50;
int mp[maxn][maxn];
int vis[maxn][maxn][2];
int n, m , k;
int cnt;
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
bool in(int x, int y){
    if(x < 1 || x > n || y < 1 || y > m) return false;
    return true;
}
void dfs(int x, int y)
{
    //cout<<"x:"<
    if(mp[x][y] == 0) return;
    mp[x][y] = 0;
    cnt++;
    for(int i = 0; i < 4; ++i){
        int xx = x + dx[i];
        int yy = y + dy[i];
        if(!in(xx, yy)) continue;
        if(i < 2) vis[xx][yy][0] --;
        else vis[xx][yy][1]--;
    }
    for(int i = 0; i < 4; ++i){
        int xx = x + dx[i];
        int yy = y + dy[i];
        if(!in(xx, yy)) continue;
        if(vis[xx][yy][0] < 2 && vis[xx][yy][1] < 2){
            dfs(xx, yy);
        }
    }
    return;
}
int main()
{
	int T;cin>>T;
	while(T--){
        scanf("%d%d%d", &n, &m, &k);
        for(int i = 1; i <= n; ++i){
            for(int j = 1; j <= m; ++j){
                mp[i][j] = 1;
                vis[i][j][0] = vis[i][j][1] = 2;
            }
        }
        while(k--){
            cnt = 0;
            int x, y;
            scanf("%d%d", &x, &y);
            dfs(x, y);
            printf("%d\n", cnt);
        }
	}
}