http://poj.org/problem?id=2195
第一个最佳匹配的题.可耻的模板都不会套。
1 #include <cstdio>
2 #include <cstring>
3 #include <cmath>
4 #include <algorithm>
5 using namespace std;
6
7 const int maxn = 2000 + 10;
8
9 const int INF = 0x7fffffff;
10
11 int n,m;
12 int W[maxn][maxn]; //存储权值
13 int Lx[maxn], Ly[maxn]; // 顶标
14 int left[maxn]; // left[i]为右边第i个点的匹配点编号
15 bool S[maxn], T[maxn]; // S[i]和T[i]为左/右第i个点是否已标记
16
17
18 bool match(int i){
19 S[i] = true;
20 for(int j = 1; j <= m; j++) if (Lx[i]+Ly[j]==W[i][j] && !T[j])
21 {
22 T[j] = true;
23 if (!left[j] || match(left[j]))
24 {
25 left[j] = i;
26 return true;
27 }
28 }
29 return false;
30 }
31
32 void update(){
33 int a = INF;
34 for(int i = 1; i <= n; i++) if(S[i])
35 for(int j = 1; j <= m; j++) if(!T[j])
36 a = min(a, W[i][j]-Lx[i]-Ly[j]); //若是最大权匹配则Lx[i]+Ly[j]-W[i][j]; 同样是取最小值
37 for(int i = 1; i <= n; i++) {
38 if(S[i]) Lx[i] -= a;
39 if(T[i]) Ly[i] += a;
40 }
41 }
42
43 void KM() {
44 for(int i = 1; i <= n; i++) {
45 left[i] = Lx[i] = Ly[i] = 0;
46 for(int j = 1; j <= m; j++)
47 Lx[i] = min(Lx[i], W[i][j]); //若是最大权匹配.则初始值顶标取最大值,最小匹配则去最小值
48 }
49 for(int i = 1; i <= n; i++) {
50 for(;;) {
51 for(int j = 1; j <= m; j++) S[j] = T[j] = 0;
52 if(match(i)) break; else update();
53 }
54 }
55 }
56 char s[305];
57 struct point
58 {
59 int x,y;
60 };
61 point man[305],house[305];
62 int main(){
63 //freopen("a.txt","r",stdin);
64 int sum,man_num,house_num;
65 while(scanf("%d%d",&n,&m)!=EOF&&(n+m))
66 {
67 int i,j;
68 man_num=1,house_num=1;
69 for(int i=1;i<=n;i++)
70 {
71 scanf("%s",s+1);
72 for(int j=1;j<=m;j++)
73 {
74 if(s[j]==‘m‘)
75 {
76 man[man_num].x=i;
77 man[man_num++].y=j;
78 }
79 else if(s[j]==‘H‘)
80 {
81 house[house_num].x=i;
82 house[house_num++].y=j;
83 }
84 }
85 }
86 man_num--,house_num--;
87 // printf("%d %d\n",man_num,house_num);
88 n=man_num;m=house_num;
89 for(int i=1;i<=man_num;i++)
90 for(int j=1;j<=house_num;j++)
91 W[i][j]=abs(man[i].x-house[j].x)+abs(man[i].y-house[j].y);
92 sum=0;
93 KM(); // 最大权匹配
94 for(i=1;i<=m;i++)
95 sum+=W[left[i]][i];
96 printf("%d\n",sum);
97 }
98 //for(int i = 1; i <= n; i++) printf("left[%d]=%d\n", i,left[i]);
99 //getch();
100 return 0;
101 }
当然还可以用费用流来做.
算出每一个到每一个房子的距离后建图。
源点与人连，容量1，费用0
人与每个房子都要连，容量1，费用为距离
每个房子与汇点连，容量1，费用0
求一次最小费用即可
另外此题数据的范围开大点。
Bellmanford:
1 #include<cstdio>
2 #include<cstring>
3 #include<queue>
4 #include<vector>
5 #include<algorithm>
6 #include<cassert>
7 using namespace std;
8
9 const int maxn = 2000 + 20;
10 const int INF = 1000000000;
11
12 struct Edge
13 {
14 int from, to, cap, flow, cost;
15 Edge(int u, int v, int c, int f, int w):from(u),to(v),cap(c),flow(f),cost(w) {}
16 };
17
18 struct MCMF
19 {
20 int n, m;
21 vector<Edge> edges;
22 vector<int> G[maxn];
23 int inq[maxn]; // 是否在队列中
24 int d[maxn]; // Bellman-Ford
25 int p[maxn]; // 上一条弧
26 int a[maxn]; // 可改进量
27
28 void init(int n)
29 {
30 this->n = n;
31 for(int i = 0; i < n; i++) G[i].clear();
32 edges.clear();
33 }
34
35 void AddEdge(int from, int to, int cap, int cost)
36 {
37 edges.push_back(Edge(from, to, cap, 0, cost));
38 edges.push_back(Edge(to, from, 0, 0, -cost));
39 m = edges.size();
40 G[from].push_back(m-2);
41 G[to].push_back(m-1);
42 }
43
44 bool BellmanFord(int s, int t, int flow_limit, int& flow, int& cost)
45 {
46 for(int i = 0; i < n; i++) d[i] = INF;
47 memset(inq, 0, sizeof(inq));
48 d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
49
50 queue<int> Q;
51 Q.push(s);
52 while(!Q.empty())
53 {
54 int u = Q.front(); Q.pop();
55 inq[u] = 0;
56 for(int i = 0; i < G[u].size(); i++)
57 {
58 Edge& e = edges[G[u][i]];
59 if(e.cap > e.flow && d[e.to] > d[u] + e.cost)
60 {
61 d[e.to] = d[u] + e.cost;
62 p[e.to] = G[u][i];
63 a[e.to] = min(a[u], e.cap - e.flow);
64 if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
65 }
66 }
67 }
68 if(d[t] == INF) return false;
69 if(flow + a[t] > flow_limit) a[t] = flow_limit - flow;
70 flow += a[t];
71 cost += d[t] * a[t];
72 for(int u = t; u != s; u = edges[p[u]].from)
73 {
74 edges[p[u]].flow += a[t];
75 edges[p[u]^1].flow -= a[t];
76 }
77 return true;
78 }
79
80 // 需要保证初始网络中没有负权圈
81 int MincostFlow(int s, int t, int flow_limit, int& cost)
82 {
83 int flow = 0; cost = 0;
84 while(flow < flow_limit && BellmanFord(s, t, flow_limit, flow, cost));
85 return flow;
86 }
87
88 };
89
90 MCMF g;
91 typedef struct{
92 int x,y;
93 }Point;
94
95 Point man[maxn],home[maxn];
96 int man_sum,home_sum;
97
98
99 int main()
100 {
101 int n,m,i,j;
102 char s[maxn];
103 while(~scanf("%d%d",&n,&m) && n || m)
104 {
105 getchar();
106 man_sum=home_sum=1;
107 for(i=1;i<=n;i++)
108 { //记录下人与房子的坐标
109 gets(s);
110 for(j=0;j<m;j++)
111 {
112 if(s[j]==‘m‘)
113 {
114 man[man_sum].x=i;
115 man[man_sum++].y=j+1;
116 }
117 if(s[j]==‘H‘)
118 {
119 home[home_sum].x=i;
120 home[home_sum++].y=j+1;
121 }
122 }
123 }
124 man_sum--; home_sum--;
125 g.init(man_sum+home_sum+2);
126
127 for(i=1;i<=man_sum;i++) g.AddEdge(0, i, 1, 0); //源点指向人
128
129 for(i=1;i<=man_sum;i++) //人指向房子
130 {
131 for(j=man_sum+1;j<=man_sum+home_sum;j++)
132 {
133 g.AddEdge(i, j, 1, abs(man[i].x-home[j-man_sum].x)+abs(man[i].y-home[j-man_sum].y));
134 }
135 }
136
137 for(i=man_sum+1;i<=man_sum+home_sum;i++) g.AddEdge(i, man_sum+home_sum+1, 1, 0); //房子指向汇点
138
139 int cost;
140 g.MincostFlow(0, man_sum+home_sum+1, man_sum, cost);
141 printf("%d\n", cost);
142 }
143 return 0;
144 }
SPFA:
参考：http://blog.csdn.net/lenleaves/article/details/7904588
1 #include<iostream>
2 #include<algorithm>
3 #include<cstring>
4 #include<string>
5 #include<queue>
6 #include<cmath>
7 #include<cstdio>
8
9 using namespace std;
10
11 #define MAXN 2000
12 #define MAXM 110000
13 #define INF 0x3FFFFFF
14
15 struct edge
16 {
17 int to,c,w,next;
18 };
19
20 edge e[MAXM];
21 bool in[MAXN];
22 int hx[500],hy[500],mx[500],my[500];
23 int head[MAXN],dis[MAXN],pre[MAXN],en,maxflow,mincost;
24 int vn,st,ed,r,c,hn,mn;
25
26 void add(int a,int b,int c,int d)
27 {
28 e[en].to=b;
29 e[en].c=c;
30 e[en].w=d;
31 e[en].next=head[a];
32 head[a]=en++;
33 e[en].to=a;
34 e[en].c=0;
35 e[en].w=-d;
36 e[en].next=head[b];
37 head[b]=en++;
38 }
39
40 bool spfa(int s)
41 {
42 queue<int> q;
43 for(int i=0;i<=vn;i++)
44 {
45 dis[i]=INF;
46 in[i]=false;
47 pre[i]=-1;
48 }
49 dis[s]=0;
50 in[s]=true;
51 q.push(s);
52 while(!q.empty())
53 {
54 int tag=q.front();
55 in[tag]=false;
56 q.pop();
57 for(int i=head[tag];i!=-1;i=e[i].next)
58 {
59 int j=e[i].to;
60 if(e[i].w+dis[tag]<dis[j] && e[i].c)
61 {
62 dis[j]=e[i].w+dis[tag];
63 pre[j]=i;
64 if(!in[j])
65 {
66 q.push(j);
67 in[j]=true;
68 }
69 }
70 }
71 }
72 if(dis[ed]==INF)
73 return false;
74 return true;
75 }
76
77 void update()
78 {
79 int flow=INF;
80 for (int i=pre[ed];i!=-1;i=pre[e[i^1].to])
81 if(e[i].c<flow) flow=e[i].c;
82 for (int i=pre[ed];i!=-1;i=pre[e[i^1].to])
83 {
84 e[i].c-=flow,e[i^1].c+=flow;
85 }
86 maxflow+=flow;
87 mincost+=flow*dis[ed];
88 }
89
90 void mincostmaxflow()
91 {
92 maxflow=0,mincost=0;
93 while(spfa(st))
94 update();
95 }
96
97 void solve()
98 {
99 char s[500];
100 hn=0,mn=0,en=0;
101 memset(head,-1,sizeof(head));
102 for(int i=1;i<=r;i++)
103 {
104 scanf("%s",s);
105 for(int j=0;j<c;j++)
106 {
107 if(s[j]==‘m‘) mx[mn]=i,my[mn++]=j+1;
108 if(s[j]==‘H‘) hx[hn]=i,hy[hn++]=j+1;
109 }
110 }
111 vn=mn+hn+5;
112 st=0,ed=mn+hn+1;
113 for(int i=0;i<mn;i++)
114 for(int j=0;j<hn;j++)
115 {
116 int dist=abs(mx[i]-hx[j])+abs(my[i]-hy[j]);
117 add(i+1,j+1+mn,1,dist);
118 }
119 for(int i=0;i<mn;i++)
120 add(st,i+1,1,0);
121 for(int i=0;i<hn;i++)
122 add(i+1+mn,ed,1,0);
123 mincostmaxflow();
124 printf("%d\n",mincost);
125 }
126
127 int main()
128 {
129 while(scanf("%d%d",&r,&c)!=EOF && r+c)
130 solve();
131 return 0;
132 }
时间： 2024-11-08 23:38:50