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HDU 2819 Swap (二分匹配+破输出)
题意:给定上一个01矩阵,让你变成一个对角全是 1 的矩阵。
析:二分匹配,把行和列看成两个集合,用匈牙利算法就可以解决,主要是在输出解,在比赛时一紧张不知道怎么输出了。
输出应该是要把 match[i] = i 这样的输出,然后再改掉后面那个,真是个大傻逼输出,气死了。。。。。
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> #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 LL LNF = 1e17; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 200 + 10; const int mod = 10000007; 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; } int match[maxn]; bool vis[maxn]; vector<int> G[maxn]; void add(int u, int v){ G[u].push_back(v); G[v].push_back(u); } 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(){ while(scanf("%d", &n) == 1){ for(int i = 0; i < n * 2; ++i) G[i].clear(); for(int i = 0; i < n; ++i) for(int j = 0; j < n; ++j){ int x; scanf("%d", &x); if(x == 1) add(i, j+n); } memset(match, -1, sizeof match); int ans = 0; n <<= 1; for(int i = 0; i < n; ++i) if(match[i] == -1){ memset(vis, 0, sizeof vis); if(dfs(i)) ++ans; } if(ans * 2 != n){ printf("-1\n"); continue; } int t = n / 2; for(int i = 0; i < t; ++i) match[i] -= t; printf("%d\n", t); for(int i = 0; i < t; ++i){ printf("C %d %d\n", i+1, match[i]+1); for(int j = i+1; j < t; ++j) if(match[j] == i){ match[j] = match[i]; break; } } } return 0; }
HDU 2819 Swap (二分匹配+破输出)
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