HDU 3722 Card Game（KM最大匹配）

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;
}