Lawn of the Dead

题意：

有一个N * M的方格，我们从(1,1)出发，只能向右走或者向下走，存在一些障碍，问有多少格子是我们所能到达的
2<=n,m,k<=1e5

题解：

所有的点减去不能到达的点的个数，就是可以到达的点的个数
有障碍的地方不能达到，而我们只能向右向下走，当某个点的左边和上边都是不可达时，该点就不可达，并会对自己的右边的点和下方的点造成影响
n,m,k<=1e5，空间很大但是地雷数目有限，我们可以从上往下逐行对每一行的地雷排序后处理，对于每个地雷，找到从自己的右上角点(x-1,y+1)开始从左往右的连续不可达区间的范围，那么x这行的这个范围也不可达，我们可以用线段树来实现区间的查询与维护。处理每一行，累加后用总点数减去即可

官方代码中，将可以到达的区域赋值1，每次找第x-1行区间内最最左侧的1，然后将第x行这段区间赋值为1，剩下0即为不可到达区域

代码：

md调了半天还是不对

---

2021/7/30 0:41
问题解决，现在代码对了，睡觉

#include<bits/stdc++.h>
#define debug(a,b) printf("%s = %d\n",a,b);
#define ls (rt<<1)
#define rs ((rt<<1)|1)
typedef long long ll;
using namespace std;
//Fe~Jozky
const ll INF_ll=1e18;
const int INF_int=0x3f3f3f3f;
inline ll read(){ll s=0,w=1ll;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')w=-1ll;ch=getchar();}while(ch>='0'&&ch<='9') s=s*10ll+((ch-'0')*1ll),ch=getchar();//s=(s<<3)+(s<<1)+(ch^48);return s*w;
}
const int maxn=3e5+9; 
int n,m,k;
vector<int>vec[maxn];
struct tree{int l,r;int lazy;int sum;
}tr[2][maxn];
void pushup(int op,int rt){tr[op][rt].sum=tr[op][ls].sum+tr[op][rs].sum;
}
void solve(int op,int rt,int val){tr[op][rt].sum=(tr[op][rt].r-tr[op][rt].l+1)*val;tr[op][rt].lazy=val;
}
void pushdown(int op,int rt){solve(op,ls,tr[op][rt].lazy);solve(op,rs,tr[op][rt].lazy);tr[op][rt].lazy=-1;
}
void build(int op,int rt,int l,int r){tr[op][rt].l=l;tr[op][rt].r=r;if(l==r){tr[op][rt].sum=0;tr[op][rt].lazy=-1;return ;}int mid=l+r>>1;build(op,ls,l,mid);build(op,rs,mid+1,r);pushup(op,rt);
}
void update(int op,int rt,int L,int R,int v){if(L>tr[op][rt].r||R<tr[op][rt].l)return ;if(L<=tr[op][rt].l&&tr[op][rt].r<=R){solve(op,rt,v);return ;}if(tr[op][rt].lazy!=-1)pushdown(op,rt);update(op,ls,L,R,v);update(op,rs,L,R,v);pushup(op,rt);
}
int query(int op,int rt,int L,int R){if(!tr[op][rt].sum)return INF_int;if(L>tr[op][rt].r||R<tr[op][rt].l)return INF_int;if(tr[op][rt].l==tr[op][rt].r)return tr[op][rt].l;if(tr[op][rt].lazy!=-1)pushdown(op,rt);if(L<=tr[op][rt].l&&tr[op][rt].r<=R){if(tr[op][ls].sum>0)return query(op,ls,L,R);else return query(op,rs,L,R);}return min(query(op,ls,L,R),query(op,rs,L,R));
}
void init(){for(int i=1;i<=n;i++)vec[i].clear();memset(tr,0,sizeof(tr));
}
int main()
{//freopen("1008.in","r",stdin);int t=read();while(t--){scanf("%d%d%d",&n,&m,&k);init();for(int i=1;i<=k;i++){int x,y;scanf("%d%d",&x,&y);vec[x].push_back(y);}build(0,1,1,m);build(1,1,1,m);update(1,1,1,1,1);int op=0;ll ans=0;for(int i=1;i<=n;i++){int L=0;sort(vec[i].begin(),vec[i].end());int pos;for(int y:vec[i]){if(y-1>=L+1){pos=query(op^1,1,L+1,y-1);if(pos!=INF_int)update(op,1,pos,y-1,1);}L=y; }if(L+1<=m){pos=query(op^1,1,L+1,m);if(pos!=INF_int)update(op,1,pos,m,1);}ans+=tr[op][1].sum;update(op^1,1,1,m,0);op^=1;}printf("%lld\n",ans);}return 0;
}

官方代码：

#include<bits/stdc++.h>
using namespace std;
#define ls (x<<1)
#define rs (x<<1|1)
const int N = 1e5 + 5;
const int inf = 0x3f3f3f3f;
vector<int>e[N];
int tr[2][N << 2], lz[2][N << 2];void push_down(int f, int x, int l, int r, int mid) {if (lz[f][x] == -1)return;tr[f][ls] = lz[f][x] * (mid - l + 1);tr[f][rs] = lz[f][x] * (r - mid);lz[f][ls] = lz[f][rs] = lz[f][x];lz[f][x] = -1;
}
void update(int f,int x, int l, int r, int L, int R, int v) {if (L <= l && R >= r) {tr[f][x] = (r - l + 1) * v;lz[f][x] = v;return;}int mid = (l + r) >> 1;push_down(f, x, l, r, mid);if (R <= mid)update(f, ls, l, mid, L, R, v);else if (L > mid)update(f, rs, mid + 1, r, L, R, v);else {update(f, ls, l, mid, L, mid, v);update(f, rs, mid + 1, r, mid + 1, R, v);}tr[f][x] = tr[f][ls] + tr[f][rs];
}
int query(int f, int x, int l, int r, int L, int R) {if (!tr[f][x])return inf;if (l == r)return l;int mid = l + r >> 1;push_down(f, x, l, r, mid);if (L <= l && R >= r) {if (tr[f][ls] > 0) return query(f, ls, l, mid, L, R);else return query(f, rs, mid + 1, r, L, R);}else {if (R <= mid)return query(f, ls, l, mid, L, R);else if (L > mid)return query(f, rs, mid + 1, r, L, R);else return min(query(f, ls, l, mid, L, mid), query(f, rs, mid + 1, r, mid + 1, R));}
}
int main() {int T;scanf("%d", &T);while (T--) {int n, m, k;scanf("%d %d %d", &n, &m, &k);for (int i = 1; i <= n; ++i)e[i].clear();for (int i = 1; i <= (m << 2); ++i) {tr[0][i] = tr[1][i] = 0;lz[0][i] = lz[1][i] = -1;}for (int i = 0; i < k; ++i) {int x, y;scanf("%d %d", &x, &y);e[x].push_back(y);}long long ans = 0;update(0, 1, 1, m, 1, 1, 1);for (int x = 1; x <= n; ++x) {int l = 0;sort(e[x].begin(), e[x].end());for (auto& y : e[x]) {if (y - 1 >= l + 1) {int pos = query((x & 1) ^ 1, 1, 1, m, l + 1, y - 1);if (pos != inf)update(x & 1, 1, 1, m, pos, y - 1, 1);}l = y;}if (l + 1 <= m) {int pos = query((x & 1) ^ 1, 1, 1, m, l + 1, m);if (pos != inf)update(x & 1, 1, 1, m, pos, m, 1);}ans += tr[x & 1][1];update((x & 1) ^ 1, 1, 1, m, 1, m, 0);}printf("%lld\n", ans);}return 0;
}