传送门
最大流比较明显了，比较麻烦的是[l,r] 需要用线段树优化建图
然后考虑只选2个的限制，我们可以原点向它连流量为 2 的边，这样它的流量就是 0/1/2
如果是1怎么办呢，我们发现可以直接它多选一个而另一个武器少选一个，对答案没有影响
然后输出方案的时候要特别注意这个限制！
#include<bits/stdc++.h>
#define N 200050
#define M 3000050
using namespace std;
const int inf = 0x3fffffff;
int read(){
int cnt = 0, f = 1; char ch = 0;
while(!isdigit(ch)){ ch = getchar(); if(ch == '-') f = -1;}
while(isdigit(ch)) cnt = cnt * 10 + (ch - '0'), ch = getchar();
return cnt * f;
}
int first[N], nxt[M], to[M], w[M], tot = 1;
void add(int x, int y, int z){
nxt[++tot] = first[x], first[x] = tot, to[tot] = y, w[tot] = z;
nxt[++tot] = first[y], first[y] = tot, to[tot] = x, w[tot] = 0;
}
int n, m, st, ed, res, pos[N], ls[N], rs[N];
int type[N];
void Build(int x, int l, int r){
if(l == r){ pos[x] = l + n; return; }
pos[x] = ++res; int mid = (l+r) >> 1;
Build(x<<1, l, mid); Build(x<<1|1, mid+1, r);
add(pos[x], pos[x<<1], inf); add(pos[x], pos[x<<1|1], inf);
}
void Modify(int x, int l, int r, int L, int R, int id){
if(L<=l && r<=R){ add(id, pos[x], 1); return;}
int mid = (l+r) >> 1;
if(L<=mid) Modify(x<<1, l, mid, L, R, id);
if(R>mid) Modify(x<<1|1, mid+1, r, L, R, id);
}
int dis[N];
bool bfs(){
queue<int> q; q.push(st);
memset(dis, -1, sizeof(dis)); dis[st] = 0;
while(!q.empty()){
int x = q.front(); q.pop();
for(int i=first[x];i;i=nxt[i]){
int t = to[i]; if(w[i] && dis[t] == -1){
dis[t] = dis[x] + 1; q.push(t);
if(t == ed) return true;
}
}
} return false;
}
int dfs(int u, int flow){
if(u == ed) return flow; int ans = 0;
for(int i=first[u];i;i=nxt[i]){
int t = to[i]; if(dis[t] == dis[u] + 1){
int delta = dfs(t, min(w[i], flow));
w[i] -= delta; w[i^1] += delta;
ans += delta; flow -= delta;
}
} return ans;
}
int dinic(){
int ans = 0; while(bfs()) ans += dfs(st, inf); return ans;
}
int match[N], c[N];
struct Node{ int a, b, c;}t[N];
void Find(int x, int &ans, int from){
if(ans) return;
if(1 <= x && x <= n){ ans = x; return;}
for(int i=first[x]; i; i=nxt[i]){
int t = to[i]; if(t != from && w[i]){
Find(t, ans, x); w[i]--; return;
}
}
}
int main(){
n = read(), m = read(); res = n + m;
Build(1, 1, m); ed = res + 1;
for(int i=1; i<=n; i++){
int op = read();
if(op == 0){
int k = read(); add(st, i, 1);
while(k--){ int x = read(); add(i, x + n, 1);}
}
if(op == 1){
int l = read(), r = read();
add(st, i, 1); Modify(1, 1, m, l, r, i);
}
if(op == 2){
int a = read(), b = read(), c = read(); t[i] = (Node){a, b, c};
add(st, i, 2); add(i, a + n, 1); add(i, b + n, 1); add(i, c + n, 1);
} type[i] = op;
}
for(int i=1; i<=m; i++) add(i + n, ed, 1);
int Maxflow = dinic();
printf("%d\n", Maxflow);
for(int i=first[ed]; i; i=nxt[i]){
if(w[i]){ int t = to[i]; Find(t, match[t], ed);
}
}
for(int i=1; i<=m; i++) c[match[i + n]]++;
for(int i=1; i<=n; i++){
if(type[i] == 2 && c[i] == 1){
if(match[t[i].a + n] != i) match[t[i].a + n] = i;
else match[t[i].b + n] = i;
}
}
for(int i=1; i<=m; i++){
if(match[i + n]) printf("%d %d\n", match[i + n], i);
} return 0;
}
