LightOJ - 1222 Gift Packing（带权二分图）

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

const int MAXNODE = 510;
typedef int Type;
const Type INF = 0x3f3f3f3f; //1e20

struct KM{
int n, m;
//邻接矩阵存储边
Type g[MAXNODE][MAXNODE];
Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];
//left[i]表示Y集的i所匹配的X集的位置，right[i]表示的是X集的i匹配的Y集的位置
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 AddEdge(int u, int v, Type val) {
g[u][v] += val;
}

bool dfs(int u) {
S[u] = true;
for (int v = 0; v < m; v++) {
//如果已经在交错树中出现过了，更新的值也不会变
if (T[v]) continue;
Type tmp = Lx[u] + Ly[v] - g[u][v];
if (!tmp) {
T[v] = true;
//如果该点还没有被匹配到，表示找到了一条增广链了，反之，继续增广下去
if (left[v] == -1 || dfs(left[v])) {
left[v] = u;
right[u] = v;
return true;
}
}
else slack[v] = min(slack[v], tmp);
}
return false;
}

void update() {
Type a = INF;
//找到Y集合的不在交错树上的点
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 j = 0; j < m; j++)
if (T[j]) Ly[j] += a;
}

Type km() {
memset(left, -1 ,sizeof(left));
memset(right, -1, sizeof(right));
memset(Ly, 0, sizeof(Ly));

//初始化的时候，Lx初始化为最大的，Ly初始化为0，这样可以保证最优
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++) {
//初始化slack，这样就能达到N^3的复杂度了
for (int j = 0; j < m; j++) slack[j] = INF;

while (1) {
memset(S, 0, sizeof(S));
memset(T, 0, sizeof(T));
if (dfs(i)) break;
update();
}
}
Type ans = 0;
for (int i = 0; i < n; i++)
ans += g[i][right[i]];
return ans;
}
}km;

int n, cas = 1;
void init() {
scanf("%d", &n);
km.init(n, n);

int val;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) {
scanf("%d", &val);
km.AddEdge(i, j, val);
}
printf("Case %d: %d\n", cas++, km.km());
}
int main() {
int test;
scanf("%d", &test);
while (test--) {
init();
}
return 0;
}