1. 过山车
题目链接：hdu - 2063
题意：求二分图的最大匹配
思路：二分图最大匹配模板题，利用匈牙利算法解决，其核心是通过不停地找增广路来增加匹配中的匹配边和匹配点，找不到增广路时，达到最大匹配
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; const int N = 1010; struct node { int to, nex; }; node edge[N * N]; int k, n, m, used[N], nex[N]; int cnt, head[N]; void add_edge(int u, int v) { edge[++cnt].to = v; edge[cnt].nex = head[u]; head[u] = cnt; } bool find(int u) { for (int i = head[u]; 0 != i; i = edge[i].nex) { int v = edge[i].to; if (!used[v]) { used[v] = 1; if (nex[v] == 0 || find(nex[v])) { nex[v] = u; return true; } } } return false; } int match() { int res = 0; for (int i = 1; i <= n; i++) { memset(used, 0, sizeof(used)); if (find(i)) res++; } return res; } int main() { while (scanf("%d", &k) && k) { cnt = 0; memset(head, 0, sizeof(head)); memset(nex, 0, sizeof(nex)); scanf("%d%d", &n, &m); while (k--) { int u, v; scanf("%d%d", &u, &v); add_edge(u, n + v); } printf("%d ", match()); } return 0; }
2. Antenna Placement
题目链接：poj - 3020
题意：在平面上有若干点，可以用一个圆圈覆盖相邻的两个点，问最少需要多少圆圈能将所有点覆盖
思路：二分图求最小路径覆盖，关键在于如何建图，可以采用黑白相间染色的办法，被染成黑色的点放在一个集合，被染成白色的点放在另一个集合，将相邻的黑色点和白色点之间连一条边，求出最大匹配数，二分图最小路径覆盖数=点的总数-二分图最大匹配数
#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; const int N = 50; const int M = 410; struct node { int nex, to; }; int T, h, w, n, m, b[N][N]; int used[M], nex[M], head[M], cnt; node edge[M * M]; char s[N][N]; void add_edge(int u, int v) { edge[++cnt].to = v; edge[cnt].nex = head[u]; head[u] = cnt; } bool find(int u) { for (int i = head[u]; 0 != i; i = edge[i].nex) { int v = edge[i].to; if (used[v]) continue; used[v] = 1; if (0 == nex[v] || find(nex[v])) { nex[v] = u; return true; } } return false; } int match() { int res = 0; for (int i = 1; i <= n; i++) { memset(used, 0, sizeof(used)); if (find(i)) res++; } return res; } void init() { cnt = 0; memset(nex, 0, sizeof(nex)); memset(head, 0, sizeof(head)); memset(b, 0, sizeof(b)); } int main() { scanf("%d", &T); while (T--) { init(); scanf("%d%d", &h, &w); for (int i = 1; i <= h; i++) scanf("%s", s[i] + 1); n = m = 0; for (int i = 1; i <= h; i++) { for (int k = 1; k <= w; k++) { if ('*' != s[i][k]) continue; if (0 == (i + k) % 2) b[i][k] = ++n; else b[i][k] = ++m; } } for (int i = 1; i <= h; i++) { for (int k = 1; k <= w; k++) { if ('*' != s[i][k] || 0 != (i + k) % 2) continue; if (b[i - 1][k]) add_edge(b[i][k], n + b[i - 1][k]); if (b[i + 1][k]) add_edge(b[i][k], n + b[i + 1][k]); if (b[i][k + 1]) add_edge(b[i][k], n + b[i][k + 1]); if (b[i][k - 1]) add_edge(b[i][k], n + b[i][k - 1]); } } int res = match(); printf("%d ", n + m - res); } return 0; }
3. Treasure Exploration
题目链接：poj - 2594
题意：给出一个由n个点m条边组成的有向无环图，求最少需要多少路径，使得这些路径可以覆盖所有点，每个点可以被多条路径覆盖
思路：二分图求最小路径覆盖，因为有的点可以被多条路径覆盖，可能存在相交路径，所以先要用floyd求出传递闭包，将求最小可相交路径转换成最小不可相交路径，然后再用二分图求最小路径覆盖，最小路径覆盖数=点的总数-二分图最大匹配数
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> using namespace std; const int N = 510; int n, m, f[N][N]; int used[N], nex[N]; void floyd() { for (int k = 1; k <= n; k++) for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) f[i][j] |= (f[i][k] & f[k][j]); } bool find(int u) { for (int i = 1; i <= n; i++) { if (!used[i] && f[u][i]) { used[i] = 1; if (0 == nex[i] || find(nex[i])) { nex[i] = u; return true; } } } return false; } int match() { int res = 0; for (int i = 1; i <= n; i++) { memset(used, 0, sizeof(used)); if (find(i)) res++; } return res; } int main() { while (scanf("%d%d", &n, &m) && 0 != (n + m)) { memset(nex, 0, sizeof(nex)); memset(f, 0, sizeof(f)); while (m--) { int u, v; scanf("%d%d", &u, &v); f[u][v] = 1; } floyd(); int res = match(); printf("%d ", n - res); } return 0; }