标签:open algorithm stream -- double cos type cstring while
题意:给定一个 n * m 的矩阵,有一些格子有目标,每次可以消灭一行或者一列,问你最少要几次才能完成。
析:把 行看成 X,把列看成是 Y,每个目标都连一条线,那么就是一个二分图的最小覆盖数,这个答案就是二分图的最大匹配,在输出解的时候,就是从匈牙利树上,从X的未盖点出发,然后标记X和Y,最后X中未标记的和Y标记的就是答案。
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> #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 double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 2000 + 10; const LL mod = 1e9 + 7; 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; } struct Edge{ int to, next; }; Edge edge[maxn*maxn/4]; int head[maxn], cnt; int match[maxn]; bool used[maxn]; bool visx[maxn], visy[maxn]; void addEdge(int u, int v){ edge[cnt].to = v; edge[cnt].next = head[u]; head[u] = cnt++; } bool dfs(int u){ used[u] = true; for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to, w = match[v]; if(w < 0 || !used[w] && dfs(w)){ match[u] = v; match[v] = u; return true; } } return false; } void dfs1(int u){ visx[u] = 1; for(int i = head[u]; ~i; i = edge[i].next){ int v = edge[i].to; if(visy[v]) continue; visy[v] = 1; dfs1(match[v]); } } int main(){ int r, c; while(scanf("%d %d %d", &r, &c, &m) == 3 && r+c+m){ ms(head, -1); cnt = 0; for(int i = 0; i < m; ++i){ int u, v; scanf("%d %d", &u, &v); --u, --v; addEdge(u, v + r); } n = r + c; ms(match, -1); int ans = 0; for(int i = 0; i < n; ++i) if(match[i] < 0){ ms(used, 0); if(dfs(i)) ++ans; } printf("%d", ans); ms(visx, 0); ms(visy, 0); ms(used, 0); for(int i = 0; i < r; ++i) if(match[i] != -1) used[i] = 1; for(int i = 0; i < r; ++i) if(!used[i]) dfs1(i); for(int i = 0; i < r; ++i) if(!visx[i]) printf(" r%d", i + 1); for(int i = r; i < n; ++i) if(visy[i]) printf(" c%d", i + 1 - r); printf("\n"); } return 0; }
标签:open algorithm stream -- double cos type cstring while
原文地址:http://www.cnblogs.com/dwtfukgv/p/7604248.html