题意:给定 n 个人成绩排名区间,然后问你最多有多少人成绩是真实的。
析:真是没想到二分匹配,。。。。后来看到,一下子就明白了,原来是水题,二分匹配,只要把每个人和他对应的区间连起来就好,跑一次二分匹配,裸的。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 101000 + 10; const int mod = 10; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r > 0 && r <= n && c > 0 && c <= m; } vector<int> G[maxn]; void add(int u, int v){ G[u].push_back(v); G[v].push_back(u); } int match[maxn]; bool vis[maxn]; bool dfs(int u){ vis[u] = true; for(int i = 0; i < G[u].size(); ++i){ int v = G[u][i]; int w = match[v]; if(w == -1 || !vis[w] && dfs(w)){ match[u] = v; match[v] = u; return true; } } return false; } int main(){ int T; cin >> T; while(T--){ scanf("%d", &n); for(int i = 0; i < maxn; ++i) G[i].clear(); for(int i = 0; i < n; ++i){ int u, v; scanf("%d %d", &u, &v); for(int j = u+100; j <= v+100; ++j) add(i, j); } memset(match, -1, sizeof match); int ans = 0; for(int i = n-1; i >= 0; --i) if(match[i] < 0){ memset(vis, 0, sizeof vis); if(dfs(i)) ++ans; } printf("%d\n", ans); int cnt = 0; for(int i = 0; i < n; ++i) if(match[i] != -1) printf("%d%c", i+1, ++cnt == ans ? '\n' : ' '); } return 0; }