标签:double iostream freopen namespace 攻击 pre ios void its
题意:给定一个 n * n的图,X是卒, . 是空位置,让你放尽量多的车,使得他们不互相攻击。
析:把每行连续的 . 看成X集体的一个点,同理也是这样,然后求一个最大匹配即可。
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> #include <assert.h> #include <bitset> #include <numeric> #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 fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) //#define sz size() #define pu push_up #define pd push_down #define cl clear() #define all 1,n,1 #define FOR(i,x,n) for(int i = (x); i < (n); ++i) #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 = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 100 + 10; const int maxm = 3e5 + 10; const ULL mod = 3; 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; } char s[maxn][maxn]; int a[maxn][maxn], b[maxn][maxn]; struct Edge{ int to, next; }; Edge edges[maxn*maxn]; int head[maxn*maxn], cnt; void addEdge(int u, int v){ edges[cnt].to = v; edges[cnt].next = head[u]; head[u] = cnt++; } int match[maxn*maxn]; bool used[maxn*maxn]; bool dfs(int u){ used[u] = 1; for(int i = head[u]; ~i; i = edges[i].next){ int v = edges[i].to, w = match[v]; if(w == -1 || !used[w] && dfs(w)){ match[u] = v; match[v] = u; return true; } } return false; } int main(){ while(scanf("%d", &n) == 1){ int idx = 0; for(int i = 0; i < n; ++i){ scanf("%s", s[i]); if(s[i][0] == ‘.‘) a[i][0] = ++idx; for(int j = 1; j < n; ++j) if(s[i][j] == ‘.‘ && s[i][j] == s[i][j-1]) a[i][j] = a[i][j-1]; else if(s[i][j] == ‘.‘) a[i][j] = ++idx; } int row = idx; for(int j = 0; j < n; ++j){ if(s[0][j] == ‘.‘) b[0][j] = ++idx; for(int i = 1; i < n; ++i) if(s[i][j] == ‘.‘ && s[i][j] == s[i-1][j]) b[i][j] = b[i-1][j]; else if(s[i][j] == ‘.‘) b[i][j] = ++idx; } cnt = 0; ms(head, -1); FOR(i, 0, n) FOR(j, 0, n) if(s[i][j] == ‘.‘){ addEdge(a[i][j], b[i][j]); addEdge(b[i][j], a[i][j]); } int ans = 0; ms(match, -1); for(int i = 1; i <= row; ++i) if(match[i] < 0){ ms(used, 0); if(dfs(i)) ++ans; } printf("%d\n", ans); } return 0; }
UVaLive 6525 Attacking rooks (二分图最大匹配)
标签:double iostream freopen namespace 攻击 pre ios void its
原文地址:http://www.cnblogs.com/dwtfukgv/p/7816717.html