关于二分图的完美匹配问题
前言
占坑
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
using std::min; using std::max;
using std::swap; using std::sort;
using std::__gcd;
typedef long long ll;
template<typename T>
void read(T &x) {
int flag = 1; x = 0; char ch = getchar();
while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
}
const int N = 3e2 + 10, Inf = 1e9 + 7;
int n, w[N][N], match[N], ret;
int lx[N], ly[N];//定点标号
bool visx[N], visy[N];
void upt0(int &x, int y) { if(x < y) x = y; }
void upt1(int &x, int y) { if(x > y) x = y; }
bool Hungary(int x) {
visx[x] = 1;
for(int y = 1; y <= n; ++y) {
if(!visy[y] && lx[x] + ly[y] == w[x][y]) {
visy[y] = true;
if(!match[y] || Hungary(match[y])) { match[y] = x; return 1; }
}
} return 0;
}
int KM() {
for(int i = 1; i <= n; ++i) {
lx[i] = -Inf, ly[i] = 0;
for(int j = 1; j <= n; ++j) upt0(lx[i], w[i][j]);
}
memset(match, 0, sizeof match);
for(int x = 1; x <= n; ++x)
while(1) {
memset(visx, 0, sizeof visx), memset(visy, 0, sizeof visy);
if(Hungary(x)) break;
int inc = Inf;//增量标记
for(int i = 1; i <= n; ++i)
if(visx[i])
for(int j = 1; j <= n; ++j)
if(!visy[j]) upt1(inc, lx[i] + ly[j] - w[i][j]);
for(int i = 1; i <= n; ++i) {
if(visx[i]) lx[i] -= inc;
if(visy[i]) ly[i] += inc;
}
}
for(int i = 1; i <= n; ++i)
if(match[i]) ret += w[match[i]][i];
return ret;
}
int main () {
while(scanf("%d", &n) != EOF) {
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j)
read(w[i][j]);
printf("%d\n", KM()), ret = 0;
}
return 0;
}