POJ-1087-A Plug for UNIX
http://poj.org/problem?id=1087
n1个插座,n2个电器及其对应的插座,n3个转化器,前一个插座可以转化为后一个插座,问最少有多少设备没有插座用,转换器数量不限
最大流,源点向插座建边,容量为1,电器向汇点建边,容量为1,相应的插座和电器连边,容量为1,前一个插座转化为后一个插座,后一个插座向前一个插座建边,容量为无穷大,求得的最大流即为最多配对的电器
#include <iostream> #include <cstdio> #include <cstring> #include <queue> #include <algorithm> using namespace std; const int maxn = 30000; const int maxm = 50000; const int INF=0x7fffffff; int idx; int cur[maxn], pre[maxn]; int dis[maxn], gap[maxn]; int aug[maxn], head[maxn]; char name[3000][30]; int k; struct Node { int u, v, w; int next; }edge[maxm]; void addEdge(int u, int v, int w) { edge[idx].u = u; edge[idx].v = v; edge[idx].w = w; edge[idx].next = head[u]; head[u] = idx++; edge[idx].u = v; edge[idx].v = u; edge[idx].w = 0; edge[idx].next = head[v]; head[v] = idx++; } int sol(char *str) { int i; if(k==0) { strcpy(name[1],str); k=1; return 1; } for(i=1;i<=k;i++) if(strcmp(name[i],str)==0) return i; k++; strcpy(name[k],str); return k; } int SAP(int s, int e, int n) { int max_flow = 0, v, u = s; int id, mindis; aug[s] =INF; pre[s] = -1; memset(dis, 0, sizeof(dis)); memset(gap, 0, sizeof(gap)); gap[0] = n; // 我觉得这一句要不要都行,因为dis[e]始终为0 for (int i = 0; i <= n; ++i) // 初始化当前弧为第一条弧 cur[i] = head[i]; while (dis[s] < n) { bool flag = false; if (u == e) { max_flow += aug[e]; for (v = pre[e]; v != -1; v = pre[v]) // 路径回溯更新残留网络 { id = cur[v]; edge[id].w -= aug[e]; edge[id^1].w += aug[e]; aug[v] -= aug[e]; // 修改可增广量,以后会用到 if (edge[id].w == 0) u = v; // 不回退到源点,仅回退到容量为0的弧的弧尾 } } for (id = cur[u]; id != -1; id = edge[id].next) { // 从当前弧开始查找允许弧 v = edge[id].v; if (edge[id].w > 0 && dis[u] == dis[v] + 1) // 找到允许弧 { flag = true; pre[v] = u; cur[u] = id; aug[v] = min(aug[u], edge[id].w); u = v; break; } } if (flag == false) { if (--gap[dis[u]] == 0) break; /* gap优化,层次树出现断层则结束算法 */ mindis = n; cur[u] = head[u]; for (id = head[u]; id != -1; id = edge[id].next) { v = edge[id].v; if (edge[id].w > 0 && dis[v] < mindis) { mindis = dis[v]; cur[u] = id; // 修改标号的同时修改当前弧 } } dis[u] = mindis + 1; gap[dis[u]]++; if (u != s) u = pre[u]; // 回溯继续寻找允许弧 } } return max_flow; } int main() { int i,n1,n2,n3; int st,ed; int a,b; char str1[30],str2[30]; while(scanf("%d",&n1)!=EOF) { idx = 0; memset(head, -1, sizeof(head)); st=0; k=0; for(i=1;i<=n1;i++) { scanf("%s",str1); a=sol(str1); addEdge(st,a,1); //源点向插座建边 } scanf("%d",&n2); ed=n1+n2+1; for(i=1;i<=n2;i++) { scanf("%s %s",str1,str2); a=sol(str1); b=sol(str2); addEdge(b,a,1); //插座向电器建立边 addEdge(a,ed,1); //电器向汇点建边 } scanf("%d",&n3); for(i=1;i<=n3;i++) { scanf("%s %s",str1,str2); a=sol(str1); b=sol(str2); addEdge(b,a,INF); } printf("%d\n",n2-SAP(st, ed, k+2)); } return 0; }