KM hdu2255 奔小康赚大钱

传送门：点击打开链接

题意：最裸的带权二分图匹配。

思路：KM算法，复杂度为O(n^3)，比费用流的效率要快一些，所以有时候必须用KM算法。

但是KM算法的适用范围比较小，要学会去分辨.

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;

const int MX = 3e2 + 5;
/* KM算法
* 复杂度O(nx*nx*ny)
* 求最大权匹配
* 若求最小权匹配，可将权值取相反数，结果取相反数
* 点的编号从1开始
*/
const int INF = 0x3f3f3f3f;
int nx, ny; //两边的点数
int G[MX][MX];//二分图描述
int linker[MX], lx[MX], ly[MX]; //y中各点匹配状态，x,y中的点标号
int slack[MX];
bool visx[MX], visy[MX];

bool DFS(int x) {
visx[x] = 1;
for(int y = 1; y <= ny; y++) {
if(visy[y]) continue;
int tmp = lx[x] + ly[y] - G[x][y];
if(tmp == 0) {
visy[y] = 1;
if(linker[y] == -1 || DFS(linker[y])) {
linker[y] = x;
return 1;
}
} else if(slack[y] > tmp) {
slack[y] = tmp;
}
}
return 0;
}
int KM() {
memset(linker, -1, sizeof(linker));
memset(ly, 0, sizeof(ly));
for(int i = 1; i <= nx; i++) {
lx[i] = -INF;
for(int j = 1; j <= ny; j++) {
if(G[i][j] > lx[i]) lx[i] = G[i][j];
}
}
for(int x = 1; x <= nx; x++) {
for(int i = 1; i <= ny; i++) slack[i] = INF;
while(true) {
memset(visx, 0, sizeof(visx));
memset(visy, 0, sizeof(visy));
if(DFS(x)) break;
int d = INF;
for(int i = 1; i <= ny; i++) {
if(!visy[i] && d > slack[i]) d = slack[i];
}
for(int i = 1; i <= nx; i++) {
if(visx[i]) lx[i] -= d;
}
for(int i = 1; i <= ny; i++) {
if(visy[i]) ly[i] += d;
else slack[i] -= d;
}
}
}
int res = 0;
for(int i = 1; i <= ny; i++) {
if(linker[i] != -1) res += G[linker[i]][i];
}
return res;
}
int main() {
int n; //FIN;
while(~scanf("%d", &n)) {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
scanf("%d", &G[i][j]);
}
}
nx = ny = n;
printf("%d\n", KM());
}
return 0;
}