CFgym102394A - Artful Paintings(差分约束+二分)
题目
思路
在队友帮助下,理清了思路。
设每个点为\(x_i\),它们的取值为0或1。设前缀和为\(S_i\)。
题目具有单调性,涂色越多越能符合条件。因此可以考虑二分。假设二分值为mid,那么约束条件如下:
- \(S_r-S_{l-1} \ge K_i\)
- \(mid - (S_r-S_{l-1}) \ge K_j\)
- \(S_i-S_{i-1} \le 1\)
- \(S_i-S_{i-1} \ge 0\)
- \(S_n-S_0 \le mid\)
- \(S_n-S_0 \ge mid\)
一定要注意把约束条件列全。
由于\(S_0\)恒等于0,所以令dist[0]=0,从0点开始跑判负环。直接跑可能会超时,如果某时刻在spfa中有dist[i]<0,说明一定有负环,因为图是双向联通。
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll maxn = 1e4 + 10;
ll vis[maxn];
bool used[maxn];
vector<pair<ll, ll> > g[maxn];
ll dist[maxn];
bool inQue[maxn];
queue<ll> que;
ll n, m1, m2;
void init(ll n) {
for(ll i = 0; i <= n; i++) {
g[i].clear();
dist[i] = INF;
vis[i] = 0;
used[i] = 0;
inQue[i] = 0;
}
}
bool spfa(ll src) {
dist[src] = 0;
while (!que.empty()) que.pop();
que.push(src);
inQue[src] = true;
while (!que.empty()) {
ll u = que.front();
used[u] = 1;
que.pop();
for (ll i = 0; i < g[u].size(); i++) {
if (dist[u] + g[u][i].second < dist[g[u][i].first]) {
dist[g[u][i].first] = dist[u] + g[u][i].second;
if(dist[g[u][i].first] < 0) return false; // 剪枝
if (!inQue[g[u][i].first]) {
inQue[g[u][i].first] = true;
vis[g[u][i].first]++;
if(vis[g[u][i].first] >= n + 1) return false;
que.push(g[u][i].first);
}
}
}
inQue[u] = false;
}
return true;
}
struct edge {
ll l, r, w;
};
vector<edge> ed;
void build(ll n, ll mid) {
for(ll i = 0; i < m1; i++) {
ll l = ed[i].l, r = ed[i].r, w = ed[i].w;
g[r].push_back({l - 1, -w});
}
for(ll i = m1; i < m1 + m2; i++) {
ll l = ed[i].l, r = ed[i].r, w = ed[i].w;
g[l - 1].push_back({r, mid - w});
}
for(ll i = 1; i <= n; i++) {
g[i].push_back({i - 1, 0});
}
for(ll i = 1; i <= n; i++) {
g[i - 1].push_back({i, 1});
}
g[n].push_back({0, -mid});
g[0].push_back({n, mid});
}
int main() {
ll t;
cin >> t;
while(t--) {
ed.clear();
cin >> n >> m1 >> m2;
for(ll i = 0; i < m1 + m2; i++) {
ll l, r, w;
cin >> l >> r >> w;
ed.push_back({l, r, w});
}
ll l = 0, r = n;
while(l <= r) {
ll mid = (l + r) / 2;
init(n + 1);
build(n, mid);
if(spfa(0)) {
r = mid - 1;
} else {
l = mid + 1;
}
}
cout << l << endl;
}
}