题意
有若干个人在数轴上跑步，每个人可能往左跑，也可能往右跑，但是方向不能变。速度是(1 m/s)
现在给定(n)个报告，每个报告给出(t)时刻在(x)位置有人经过。
问至少有多少个人在跑步。
思路
- 注：坐标旋转公式：
[egin{pmatrix}
x \
y
end{pmatrix}
=
egin{pmatrix}
1 & 1 \
1 & -1
end{pmatrix}
egin{pmatrix}
a \
b
end{pmatrix}
]
即，
[egin{cases}
x = a + b, &\
y = a - b, &
end{cases}
]
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
typedef pair<int, int> pii;
const int N = 200010, M = 4 * N, inf = 1e8;
int n, S, T;
int h[N], e[M], ne[M], f[M], idx;
int cur[N], d[N];
vector<int> nums;
pii report[N];
void add(int a, int b, int c)
{
e[idx] = b, f[idx] = c, ne[idx] = h[a], h[a] = idx ++;
e[idx] = a, f[idx] = 0, ne[idx] = h[b], h[b] = idx ++;
}
bool bfs()
{
memset(d, -1, sizeof(d));
queue<int> que;
que.push(S);
d[S] = 0, cur[S] = h[S];
while(que.size()) {
int t = que.front();
que.pop();
for(int i = h[t]; ~i; i = ne[i]) {
int ver = e[i];
if(d[ver] == -1 && f[i]) {
d[ver] = d[t] + 1;
cur[ver] = h[ver];
if(ver == T) return true;
que.push(ver);
}
}
}
return false;
}
int find(int u, int limit)
{
if(u == T) return limit;
int flow = 0;
for(int i = cur[u]; ~i && flow < limit; i = ne[i]) {
cur[u] = i;
int ver = e[i];
if(d[ver] == d[u] + 1 && f[i]) {
int t = find(ver, min(f[i], limit - flow));
if(!t) d[ver] = -1;
f[i] -= t, f[i ^ 1] += t, flow += t;
}
}
return flow;
}
int dinic()
{
int res = 0, flow;
while(bfs()) {
while(flow = find(S, inf)) {
res += flow;
}
}
return res;
}
int find(int x)
{
return lower_bound(nums.begin(), nums.end(), x) - nums.begin() + 1;
}
int main()
{
int TT;
scanf("%d", &TT);
while(TT --) {
scanf("%d", &n);
memset(h, -1, sizeof(h));
idx = 0;
nums.clear();
for(int i = 1; i <= n; i ++) {
int t, x;
scanf("%d%d", &t, &x);
report[i] = {t + x, t - x};
nums.push_back(t + x);
nums.push_back(t - x);
}
sort(nums.begin(), nums.end());
nums.erase(unique(nums.begin(), nums.end()), nums.end());
int m = nums.size();
S = 0, T = 2 * m + 1;
for(int i = 1; i <= m; i ++) {
add(S, i, 1);
add(m + i, T, 1);
}
for(int i = 1; i <= n; i ++) {
int a = find(report[i].first), b = find(report[i].second);
add(a, m + b, inf);
}
printf("%d
", dinic());
}
return 0;
}