HDU 3315 My Brute（KM最大匹配）

HDU 3315 My Brute

题目链接

和HDU2835是一样的思路，利用把数字离散掉来多判断一个优先级

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

const int MAXNODE = 105;

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;
}

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] == i) ans += (g[i][right[i]] - 1) / (n + 1);
else ans += g[i][right[i]] / (n + 1);
}
return ans;
}
} gao;

const int N = 105;

int n, val
, h1
, h2
, a1
, a2
;

int main() {
while (~scanf("%d", &n) && n) {
for (int i = 0; i < n; i++) scanf("%d", &val[i]);
for (int i = 0; i < n; i++) scanf("%d", &h1[i]);
for (int i = 0; i < n; i++) scanf("%d", &h2[i]);
for (int i = 0; i < n; i++) scanf("%d", &a1[i]);
for (int i = 0; i < n; i++) scanf("%d", &a2[i]);
gao.init(n, n);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int t1 = h1[i] / a2[j] + (h1[i] % a2[j] != 0);
int t2 = h2[j] / a1[i] + (h2[j] % a1[i] != 0);
if (t1 >= t2) gao.add_Edge(i, j, val[i] * (n + 1));
else gao.add_Edge(i, j, -val[i] * (n + 1));
if (i == j) gao.g[i][j]++;
}
}
int ans = gao.km();
if (ans <= 0) printf("Oh, I lose my dear seaco!\n");
else {
int cnt = 0;
for (int i = 0; i < gao.n; i++)
if (gao.right[i] == i) cnt++;
printf("%d %.3lf%%\n", ans, cnt * 1.0 / n * 100);
}
}
return 0;
}