HDU 6598 Harmonious Army 网络流建图

Harmonious Army Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 221    Accepted Submission(s): 89   Problem Description Now, Bob is playing an interesting game in which he is a general of a harmonious army. There are n soldiers in this army. Each soldier should be in one of the two occupations, Mage or Warrior. There are m pairs of soldiers having combination ability. There are three kinds of combination ability. If the two soldiers in a pair are both Warriors, the army power would be increased by a . If the two soldiers in a pair are both Mages, the army power would be increased by c . Otherwise the army power would be increased by b , and b=a/4+c/3 , guaranteed that 4|a and 3|c . Your task is to output the maximum power Bob can increase by arranging the soldiers' occupations. Note that the symbol a|b means that a divides b , e.g. , 3|12 and 8|24 .     Input There are multiple test cases. Each case starts with a line containing two positive integers n(n≤500) and m(m≤104) . In the following m lines, each line contains five positive integers u,v,a,b,c (1≤u,v≤n,u≠v,1≤a,c≤4×106,b=a/4+c/3) , denoting soldiers u and v have combination ability, guaranteed that the pair (u,v) would not appear more than once. It is guaranteed that the sum of n in all test cases is no larger than 5×103 , and the sum of m in all test cases is no larger than 5×104 .     Output For each test case, output one line containing the maximum power Bob can increase by arranging the soldiers' occupations.     Sample Input   3 2 1 2 8 3 3 2 3 4 3 6     Sample Output   12     Source 2019 Multi-University Training Contest 2     Recommend liuyiding   |   We have carefully selected several similar problems for you:  6602 6601 6600 6599 6598     题意：     n个士兵  m 个组合   每个士兵可以转职战士或者魔法师   u  和  v  组合  两个战士战力+a   两个法师战力+c  其他战力+b     求最多加多少战力   题解很详细   化方案的权值为图上边权 利用最小割模型求解   #include using namespace std; typedef long long ll; const ll maxn=505; const ll maxm=5e5+12; const ll inf=1e15; struct Edge { ll to,nxt; ll cap,flow; }edge[maxm]; ll tol; ll head[maxn]; ll Q[maxn]; ll dep[maxn],cur[maxn],sta[maxn]; void init() { tol=0; memset(head,-1,sizeof(head)); memset(dep,0,sizeof(dep)); memset(cur,0,sizeof(cur)); memset(sta,0,sizeof(sta)); memset(edge,0,sizeof(edge)); } void AddEdge(ll u,ll v,ll w,ll rw=0) { edge[tol].to=v; edge[tol].cap=w; edge[tol].flow=0; edge[tol].nxt=head[u]; head[u]=tol++; edge[tol].to=u; edge[tol].cap=rw; edge[tol].flow=0; edge[tol].nxt=head[v]; head[v]=tol++; } bool bfs(ll s,ll t,ll n) { ll front=0,tail=0; memset(dep,-1,sizeof(dep)); dep[s]=0; Q[tail++]=s; while(frontedge[i].flow&&dep[v]==-1) { dep[v]=dep[u]+1; if(v==t) return true; Q[tail++]=v; } } } return false; } ll dinic(ll s,ll t,ll n) { ll maxflow=0; while(bfs(s,t,n)) { for(ll i=0;i=0;i--) { tp=min(tp,edge[sta[i]].cap-edge[sta[i]].flow); } maxflow+=tp; for(ll i=tail-1;i>=0;i--) { edge[sta[i]].flow+=tp; edge[sta[i]^1].flow-=tp; if(edge[sta[i]].cap-edge[sta[i]].flow==0) tail=i; } u=edge[sta[tail]^1].to; } else if(cur[u]!=-1&&edge[cur[u]].cap>edge[cur[u]].flow&&dep[u]+1==dep[edge[cur[u]].to]) { sta[tail++]=cur[u]; u=edge[cur[u]].to; } else { while(u!=s&&cur[u]==-1) u=edge[sta[--tail]^1].to; cur[u]=edge[cur[u]].nxt; } } } return maxflow; } int main() { ll m,n; while(~scanf("%lld%lld",&n,&m)) { init(); ll u,v,a,b,c; ll A,C,E; ll s=0,t=n+1; ll sum=0; for(int cas=1;cas<=m;cas++) { scanf("%lld%lld%lld%lld%lld",&u,&v,&a,&b,&c); a*=2; b*=2; c*=2; sum+=a+b+c; A=(a+b)/2; C=(c+b)/2; E=-b+(a+c)/2; AddEdge(s,u,A); AddEdge(s,v,A); AddEdge(u,t,C); AddEdge(v,t,C); AddEdge(u,v,E); AddEdge(v,u,E); } ll ans=dinic(s,t,n+2); printf("%lld\n",(sum-ans)/2); } }