HDU 3722 Card Game（KM最大匹配）

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HDU 3722 Card Game

题目链接

题意：给定一些字符串，每次可以选两个a，b出来，a的前缀和b的后缀的最长公共长度就是获得的值，字符串不能重复选，问最大能获得多少值

思路：KM最大匹配，两两串建边，跑最大匹配即可

代码：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;const int MAXNODE = 205;typedef int Type;const Type INF = 0x3f3f3f3f;struct KM {int n, m;Type g[MAXNODE][MAXNODE];Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];int left[MAXNODE], right[MAXNODE];bool S[MAXNODE], T[MAXNODE];void init(int n, int m) {this->n = n;this->m = m;memset(g, 0, sizeof(g));}void add_Edge(int u, int v, Type val) {g[u][v] = val;}bool dfs(int i) {S[i] = true;for (int j = 0; j < m; j++) {if (T[j]) continue;Type tmp = Lx[i] + Ly[j] - g[i][j];if (!tmp) {T[j] = true;if (left[j] == -1 || dfs(left[j])) {left[j] = i;right[i] = j;return true;}} else slack[j] = min(slack[j], tmp);}return false;}void update() {Type a = INF;for (int i = 0; i < m; i++)if (!T[i]) a = min(a, slack[i]);for (int i = 0; i < n; i++)if (S[i]) Lx[i] -= a;for (int i = 0; i < m; i++)if (T[i]) Ly[i] += a;}Type km() {memset(left, -1, sizeof(left));memset(right, -1, sizeof(right));memset(Ly, 0, sizeof(Ly));for (int i = 0; i < n; i++) {Lx[i] = -INF;for (int j = 0; j < m; j++)Lx[i] = max(Lx[i], g[i][j]);}for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) slack[j] = INF;while (1) {memset(S, false, sizeof(S));memset(T, false, sizeof(T));if (dfs(i)) break;else update();}}Type ans = 0;for (int i = 0; i < n; i++) {if (right[i] == -1) return -1;if (g[i][right[i]] == -INF) return -1;ans += g[i][right[i]];}return ans;}} gao;int n;char str[205][1005];int cal(char *a, char *b) {int an = strlen(a), bn = strlen(b);int tot = min(an, bn);int cnt = 0;for (int i = 0; i < tot; i++) {if (a[i] != b[bn - i - 1]) break;cnt++;}return cnt;}int main() {while (~scanf("%d", &n)) {gao.init(n, n);for (int i = 0; i < n; i++) {scanf("%s", str[i]);for (int j = 0; j < i; j++) {gao.add_Edge(i, j, cal(str[i], str[j]));gao.add_Edge(j, i, cal(str[j], str[i]));}}printf("%d\n", gao.km());}return 0;}

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