HDU 3435 A new Graph Game（KM完美匹配）

HDU 3435 A new Graph Game

题目链接

题意：又是这类求环总和的最小值，一个点只能在一个环上

思路：KM完美匹配求解，不过这题点有1000，理论上n^3算法是过不了的，不过也没有更好的方法了，用费用流来做的话也是n^3，而且常数还更大

代码：

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

const int MAXNODE = 1005;

typedef int Type;
const Type INF = 0x3f3f3f3f;

struct KM {
int n;
Type g[MAXNODE][MAXNODE];
Type Lx[MAXNODE], Ly[MAXNODE], slack[MAXNODE];
int left[MAXNODE];
bool S[MAXNODE], T[MAXNODE];

void init(int n) {
this->n = n;
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
g[i][j] = -INF;
}

void add_Edge(int u, int v, Type val) {
g[u][v] = max(g[u][v], val);
}

bool dfs(int i) {
S[i] = true;
for (int j = 0; j < n; 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;
return true;
}
} else slack[j] = min(slack[j], tmp);
}
return false;
}

void update() {
Type a = INF;
for (int i = 0; i < n; i++)
if (!T[i]) a = min(a, slack[i]);
for (int i = 0; i < n; i++) {
if (S[i]) Lx[i] -= a;
if (T[i]) Ly[i] += a;
}
}

void km() {
for (int i = 0; i < n; i++) {
left[i] = -1;
Lx[i] = -INF; Ly[i] = 0;
for (int j = 0; j < n; j++)
Lx[i] = max(Lx[i], g[i][j]);
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) slack[j] = INF;
while (1) {
for (int j = 0; j < n; j++) S[j] = T[j] = false;
if (dfs(i)) break;
else update();
}
}
int ans = 0;
for (int i = 0; i < n; i++) {
if (g[left[i]][i] == -INF) {
printf("NO\n");
return;
}
ans += g[left[i]][i];
}
printf("%d\n", -ans);
}
} gao;

int t, n, m;

int main() {
int cas = 0;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
gao.init(n);
int u, v, w;
while (m--) {
scanf("%d%d%d", &u, &v, &w);
u--; v--;
gao.add_Edge(u, v, -w);
gao.add_Edge(v, u, -w);
}
printf("Case %d: ", ++cas);
gao.km();
}
return 0;
}