poj 3084 最小割
题目链接：http://poj.org/problem?id=3084
本题主要在构图上，我采用的是把要保护的房间与源点相连，有intruder的与汇点相连，相对麻烦。
#include <cstdio> #include <cmath> #include <algorithm> #include <iostream> #include <cstring> #include <queue> #include <vector> #define maxn 30 #define maxe 5000 using namespace std; const int INF = 0x3f3f3f; struct Edge{ int from,to,cap,flow; int next; }; struct Dinic{ int s,t; int head[maxn]; int cur[maxn]; Edge edges[maxe]; int d[maxn]; bool vis[maxn]; int cnt; void init(){ memset(head,-1,sizeof(head)); cnt = 0; } void addedge(int from,int to,int cap){ edges[cnt].from = from; edges[cnt].to = to; edges[cnt].cap = cap; edges[cnt].flow = 0 ; edges[cnt].next = head[from]; head[from] = cnt++; edges[cnt].from = to ; edges[cnt].to = from; edges[cnt].cap = 0; edges[cnt].flow = 0 ; edges[cnt].next = head[to]; head[to] = cnt++; } bool bfs(){ memset(vis,0,sizeof(vis)); queue<int> Q; Q.push(s); vis[s] = true; d[s] = 0; while(!Q.empty()){ int u = Q.front(); Q.pop(); for(int i=head[u];i!=-1;i=edges[i].next){ Edge& e = edges[i]; if(!vis[e.to] && e.cap>e.flow){ vis[e.to] = true; d[e.to] = d[e.from] + 1; Q.push(e.to); } } } return vis[t]; } int dfs(int u,int res){ if( u == t || res == 0) return res; int flow = 0,f; for(int& i=cur[u];i!=-1;i=edges[i].next){ //还不是很理解到cur[]的作用； Edge& e = edges[i]; if(d[e.to] == d[e.from] + 1 && (f = dfs(e.to,min(res,e.cap-e.flow)))>0){ e.flow += f; edges[i^1].flow -= f; flow += f; res -= f; if(res == 0) break; } } return flow; } int Maxflow(int S,int T){ s = S; t = T; int flow = 0; while(bfs()){ for(int i=s;i<=t;i++) cur[i] = head[i]; flow += dfs(s,INF); } return flow; } }solver; int main() { //if(freopen("input.txt","r",stdin)== NULL) {printf("Error\n"); exit(0);} int T; cin>>T; while(T--){ solver.init(); int m,n; scanf("%d%d",&m,&n); n++; int s,t; s = 0; t = m+1; solver.addedge(s,n,INF); for(int i=1;i<=m;i++){ char ch[4]; int adjnum; scanf("%s%d",ch,&adjnum); if(ch[0] == 'I'){ //printf("i %d\n",i); solver.addedge(i,t,INF); for(int j=1;j<=adjnum;j++){ int a; scanf("%d",&a); a++; solver.addedge(a,t,INF); //能从intruder所在房间到达的房间要与汇点相连 } } else{ for(int j=1;j<=adjnum;j++){ int a; scanf("%d",&a); a++; if(a == n) solver.addedge(s,i,INF); //能到达保护房间也要与源点相连； solver.addedge(i,a,1); solver.addedge(a,i,INF); //这是一直WA的地方； } } } int ans = solver.Maxflow(s,t); if(ans >= INF) printf("PANIC ROOM BREACH\n"); else printf("%d\n",ans); } }
另一种构图，简单多了
#include <cstdio> #include <cmath> #include <algorithm> #include <iostream> #include <cstring> #include <queue> #include <vector> #define maxn 30 #define maxe 5000 using namespace std; const int INF = 0x3f3f3f; struct Edge{ int from,to,cap,flow; int next; }; struct Dinic{ int s,t; int head[maxn]; int cur[maxn]; Edge edges[maxe]; int d[maxn]; bool vis[maxn]; int cnt; void init(){ memset(head,-1,sizeof(head)); cnt = 0; } void addedge(int from,int to,int cap){ edges[cnt].from = from; edges[cnt].to = to; edges[cnt].cap = cap; edges[cnt].flow = 0 ; edges[cnt].next = head[from]; head[from] = cnt++; edges[cnt].from = to ; edges[cnt].to = from; edges[cnt].cap = 0; edges[cnt].flow = 0 ; edges[cnt].next = head[to]; head[to] = cnt++; } bool bfs(){ memset(vis,0,sizeof(vis)); queue<int> Q; Q.push(s); vis[s] = true; d[s] = 0; while(!Q.empty()){ int u = Q.front(); Q.pop(); for(int i=head[u];i!=-1;i=edges[i].next){ Edge& e = edges[i]; if(!vis[e.to] && e.cap>e.flow){ vis[e.to] = true; d[e.to] = d[e.from] + 1; Q.push(e.to); } } } return vis[t]; } int dfs(int u,int res){ if( u == t || res == 0) return res; int flow = 0,f; for(int& i=cur[u];i!=-1;i=edges[i].next){ //还不是很理解到cur[]的作用； Edge& e = edges[i]; if(d[e.to] == d[e.from] + 1 && (f = dfs(e.to,min(res,e.cap-e.flow)))>0){ e.flow += f; edges[i^1].flow -= f; flow += f; res -= f; if(res == 0) break; } } return flow; } int Maxflow(int S,int T){ s = S; t = T; int flow = 0; while(bfs()){ for(int i=s;i<=t;i++) cur[i] = head[i]; flow += dfs(s,INF); } return flow; } }solver; int main() { //if(freopen("input.txt","r",stdin)== NULL) {printf("Error\n"); exit(0);} int T; cin>>T; while(T--){ solver.init(); int m,n; scanf("%d%d",&m,&n); n++; int s,t; s = 0; t = m+1; solver.addedge(n,t,INF); for(int i=1;i<=m;i++){ char ch[4]; int adjnum; scanf("%s%d",ch,&adjnum); if(ch[0] == 'I'){ solver.addedge(s,i,INF); } for(int j=1;j<=adjnum;j++){ int a; scanf("%d",&a); a++; solver.addedge(i,a,INF); solver.addedge(a,i,1); } } int ans = solver.Maxflow(s,t); if(ans >= INF) printf("PANIC ROOM BREACH\n"); else printf("%d\n",ans); } }
