E、room （费用流）

E、room （费用流）

Nowcoder University has 4n students and n dormitories ( Four students per dormitory). Students numbered from 1 to 4n. And in the first year, the i-th dormitory 's students are (x1[i],x2[i],x3[i],x4[i]), now in the second year, Students need to decide who to live with. In the second year, you get n tables such as (y1,y2,y3,y4) denote these four students want to live together. Now you need to decide which dormitory everyone lives in to minimize the number of students who change dormitory.    输入描述: The first line has one integer n.    Then there are n lines, each line has four integers (x1,x2,x3,x4) denote these four students live together in the first year    Then there are n lines, each line has four integers (y1,y2,y3,y4) denote these four students want to live together in the second year    输出描述: Output the least number of students need to change dormitory.    备注: 1<=n<=100    1<=x1,x2,x3,x4,y1,y2,y3,y4<=4n    It's guaranteed that no student will live in more than one dormitories.    示例 1 输入 2 1 2 3 4 5 6 7 8 4 6 7 8 1 2 3 5 输出 2    说明 Just swap 4 and 5

第一年的向第二年的连边，流量为1，费用为 - 第一年宿舍与第二年宿舍相同人数。

S与第一年连边，流量为1，费用为0.  第二年向T连边，流量为1，费用为0.

这样的话，跑一边费用流，求出的不用移动的学生的最大人数（负的），然后用总数相加即可。

代码：

#include #include #include #include #define clear(A, X) memset(A, X, sizeof A)#define min(A, B) ((A) < (B) ? (A) : (B))using namespace std;const int maxE = 2000000;const int maxN = 256;const int oo = 0x3f3f3f3f;struct Edge{ int v, c, w, n; Edge(){} Edge(int V, int C, int W, int N) : v(V), c(C), w(W), n(N){}};struct Node{ int l, r, w;};struct poin{ int x,y;}x1[300],x2[300];Edge edge[maxE];Node sche[maxN];int adj[maxN], cntE;int Q[maxE], head, tail;int d[maxN], inq[maxN], cur[maxN], f[maxN];int cost, flow, s, t;int N, M, K;void addedge(int u, int v, int c, int w){ edge[cntE] = Edge(v, c, w, adj[u]); adj[u] = cntE++; edge[cntE] = Edge(u, 0, -w, adj[v]); adj[v] = cntE++;}int spfa(){ f[s] = oo; clear(d, oo); clear(inq, 0); cur[s] = -1; d[s] = 0; head = tail = 0; Q[tail++] = s; inq[s] = 1; while(head != tail){ int u = Q[head++]; inq[u] = 0; for(int i = adj[u]; ~i; i = edge[i].n){ int v = edge[i].v, c = edge[i].c, w = edge[i].w; if(c && d[v] > d[u] + w){ d[v] = d[u] + w; f[v] = min(f[u], c); cur[v] = i; if(!inq[v]){ Q[tail++] = v; inq[v] = 1; } } } } if(d[t] == oo) return 0; flow += f[t]; cost += d[t] * f[t]; for(int i = cur[t]; ~i; i = cur[edge[i ^ 1].v]){ edge[i].c -= f[t]; edge[i ^ 1].c += f[t]; } return 1;}int MCMF(){ flow = cost = 0; while(spfa()); return cost;}void init(){ clear(adj, -1); cntE = 0;}int a[110][4];int b[110][4];int get(int x, int y) { int res = 0; for(int i=0;i<4;i++) for(int j=0;j<4;j++) if (a[x][i] == b[y][j]) res++; return res;}int S,T;int main(){ int n; cin>>n; for(int i=1;i<=n;i++)for(int j=0;j<4;j++)scanf("%d",&a[i][j]); for(int i=1;i<=n;i++)for(int j=0;j<4;j++)scanf("%d",&b[i][j]); s=0;t=2*n+1; init(); for(int i=1;i<=n;i++) { addedge(s, i, 1, 0); addedge(n + i, t, 1, 0); } for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) addedge(i, n + j, 1, -get(i, j)); printf("%d\n",4*n+MCMF()); return 0;}/*21 2 3 45 6 7 84 6 7 81 2 3 5*/

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