题意
Sol
我是这样考虑的:从大到小考虑每个\(l, r\),最大的\(l\)应该和最大的\(r\)匹配(不然就亏了),其次次大的\(r\)应该和次大的\(l\)匹配
然后就过了。。
/**/#include#define Pair pair #define MP(x, y) make_pair(x, y)#define fi first#define se second#define int long long #define LL long long #define rg register #define pt(x) printf("%d ", x);#define Fin(x) {freopen(#x".in","r",stdin);}#define Fout(x) {freopen(#x".out","w",stdout);}#define chmin(x, y) (x = x < y ? x : y)using namespace std;const int MAXN = 2001, INF = 1e18 + 10, mod = 1e9 + 7;const double eps = 1e-9;inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f;}int N, l[MAXN], r[MAXN], ans;priority_queue p, q;main() { N = read(); for(int i = 1; i <= N; i++) p.push(read()), q.push(read()); for(int i = 1; i <= N; i++) { int x = p.top(), y = q.top(); p.pop(); q.pop(); ans += max(x, y); } cout << ans + N; return 0; }