POJ 2195 Going Home 最小费用最大流 or KM算法
题目大意是一张地图中，有n个人要走回n个房子里，然后人只能横着或竖着走一格，求他们回家的距离总和最短。
这道题看起来是个最优匹配问题，用KM或者最小费用最大流做
首先把每个人与每个房子之间的距离求出来
然后就是套用模板了
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
struct point
{
int x, y;
}p[105], h[105];
int m, n;
int d[105][105];
char s[105][105];
int tot = 0;
class mincost
{
private:
const static int V = 205; //注意点的个数， 应该为人+房间+2， 所以至少要开到202
const static int E = 25015;
const static int INF = -1u >> 1;
struct Edge
{
int v, cap, cost;
Edge *next;
} pool[E], *g[V], *pp, *pree[V];
int T, S, dis[V], pre[V];
int n, m, flow, cirq[V];
void SPFA();
inline void addedge(int i, int j, int cap, int cost);
public:
bool initialize(int x, int y);
void mincost_maxflow();
};
void mincost::mincost_maxflow()
{
while (true)
{
SPFA();
if (dis[T] == INF)
break;
int minn = INF;
for (int i = T; i != S; i = pre[i])
minn = min(minn, pree[i]->cap);
for (int i = T; i != S; i = pre[i])
{
pree[i]->cap -= minn;
pool[(pree[i] - pool)^0x1].cap += minn;
flow += minn * pree[i]->cost;
}
tot += minn; //流量计算
}
printf("%d\n", flow);
}
void mincost::SPFA()
{
bool vst[V] = {false};
int tail = 0, u;
fill(dis,dis + n,0x7fffffff);
cirq[0] = S;
vst[S] = 1;
dis[S] = 0;
for (int i = 0; i <= tail; i++)
{
int v = cirq[i % n];
for (Edge *i = g[v]; i != NULL; i = i->next)
{
if (!i->cap)
continue;
u = i->v;
if (i->cost + dis[v] < dis[u])
{
dis[u] = i->cost + dis[v];
pree[u] = i;
pre[u] = v;
if (!vst[u])
{
tail++;
cirq[tail % n] = u;
vst[u] = true;
}
}
}
vst[v] = false;
}
}
void mincost::addedge(int i, int j, int cap, int cost)
{
pp->cap = cap;
pp->v = j;
pp->cost = cost;
pp->next = g[i];
g[i] = pp++;
}
bool mincost::initialize(int x, int y)
{
memset(g, 0, sizeof (g));
pp = pool;
n = x + y + 2; //n即为顶点的个数
S = 0;
T = x + y + 1;
for(int i = 1; i <= x; i++)
{
for(int j = 1; j <= y; j++)
{
addedge(i, x + j, 1, d[i][j]);
addedge(x + j, i, 0, -d[i][j]);
}
}
for(int i = 1; i <= x; i++)
{
addedge(0, i, 1, 0);
addedge(i, 0, 0, 0);
}
for(int i = 1; i <= y; i++)
{
addedge(x + i, T, 1, 0);
addedge(T, x + i, 0, 0);
}
flow = 0;
return true;
}
mincost g;
int main()
{
while(scanf("%d%d", &m, &n) != EOF)
{
if(m == 0 && n == 0) break;
for(int i = 0; i < m; i++)
scanf("%s", s[i]);
int hcnt = 0, pcnt = 0;
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
if(s[i][j] == 'H')
{
hcnt++;
h[hcnt].x = i;
h[hcnt].y = j;
}
else if(s[i][j] == 'm')
{
pcnt++;
p[pcnt].x = i;
p[pcnt].y = j;
}
}
}
for(int i = 1; i <= pcnt; i++)
{
for(int j = 1; j <= hcnt; j++)
{
d[i][j] = abs(p[i].x - h[j].x) + abs(p[i].y - h[j].y);
}
}
g.initialize(pcnt, hcnt);
tot = 0;
g.mincost_maxflow();
}
return 0;
}
KM算法版本， 用的DD神牛的模板http://cuitianyi.com/blog/%E6%B1%82%E6%9C%80%E5%A4%A7%E6%9D%83%E4%BA%8C%E5%88%86%E5%8C%B9%E9%85%8D%E7%9A%84km%E7%AE%97%E6%B3%95/
另外 芳哥blog里有篇介绍http://blog.csdn.net/wsniyufang/article/details/6759628
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#define MAXN 105
#define MAXM 555555
#define INF 1000000000
using namespace std;
char mp[MAXN][MAXN];
int n, m, ny, nx;
int w[MAXN][MAXN];
int lx[MAXN], ly[MAXN];
int linky[MAXN];
int visx[MAXN], visy[MAXN];
int slack[MAXN];
struct P
{
int x, y;
}house[MAXN], man[MAXN];
bool find(int x)
{
visx[x] = 1;
for(int y = 1; y <= ny; y++)
{
if(visy[y]) continue;
int t = lx[x] + ly[y] - w[x][y];
if(t == 0)
{
visy[y] = 1;
if(linky[y] == -1 || find(linky[y]))
{
linky[y] = x;
return true;
}
}
else if(slack[y] > t) slack[y] = t;
}
return false;
}
void KM()
{
memset(linky, -1, sizeof(linky));
for(int i = 1; i <= nx; i++) lx[i] = -INF;
memset(ly, 0, sizeof(ly));
for(int i = 1; i <= nx; i++)
for(int j = 1; j <= ny; j++)
if(w[i][j] > lx[i]) lx[i] = w[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(find(x)) break;
int d = INF;
for(int i = 1; i <= ny; i++)
if(!visy[i]) d = min(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 main()
{
while(scanf("%d%d", &n, &m) != EOF)
{
if(n == 0 && m == 0) break;
ny = 0, nx = 0;
for(int i = 0; i < n; i++) scanf("%s", mp[i]);
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(mp[i][j] == 'H') {++ny; house[ny].x = i, house[ny].y = j;}
else if(mp[i][j] == 'm') {++nx; man[nx].x = i, man[nx].y = j;}
for(int i = 1; i <= nx; i++)
for(int j = 1; j <= ny; j++)
w[i][j] = -(abs(house[j].x - man[i].x) + abs(house[j].y - man[i].y));
KM();
int ans = 0;
for(int i = 1; i <= ny; i++) ans += w[linky[i]][i];
printf("%d\n", -ans);
}
return 0;
}