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题目大概说一张无向图,各个结点初始有ai人,现在每个人可以选择停留在原地或者移动到相邻的结点,问能否使各个结点的人数变为bi人。
如此建容量网络:
跑最大流看看最大流是否等于∑bi。另外,注意∑ai不等于∑bi的情况。
1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 #include<algorithm> 5 using namespace std; 6 #define INF (1<<30) 7 #define MAXN 222 8 #define MAXM 222*222*2 9 10 struct Edge{ 11 int v,cap,flow,next; 12 }edge[MAXM]; 13 int vs,vt,NE,NV; 14 int head[MAXN]; 15 16 void addEdge(int u,int v,int cap){ 17 edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0; 18 edge[NE].next=head[u]; head[u]=NE++; 19 edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0; 20 edge[NE].next=head[v]; head[v]=NE++; 21 } 22 23 int level[MAXN]; 24 int gap[MAXN]; 25 void bfs(){ 26 memset(level,-1,sizeof(level)); 27 memset(gap,0,sizeof(gap)); 28 level[vt]=0; 29 gap[level[vt]]++; 30 queue<int> que; 31 que.push(vt); 32 while(!que.empty()){ 33 int u=que.front(); que.pop(); 34 for(int i=head[u]; i!=-1; i=edge[i].next){ 35 int v=edge[i].v; 36 if(level[v]!=-1) continue; 37 level[v]=level[u]+1; 38 gap[level[v]]++; 39 que.push(v); 40 } 41 } 42 } 43 44 int pre[MAXN]; 45 int cur[MAXN]; 46 int ISAP(){ 47 bfs(); 48 memset(pre,-1,sizeof(pre)); 49 memcpy(cur,head,sizeof(head)); 50 int u=pre[vs]=vs,flow=0,aug=INF; 51 gap[0]=NV; 52 while(level[vs]<NV){ 53 bool flag=false; 54 for(int &i=cur[u]; i!=-1; i=edge[i].next){ 55 int v=edge[i].v; 56 if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){ 57 flag=true; 58 pre[v]=u; 59 u=v; 60 //aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap)); 61 aug=min(aug,edge[i].cap-edge[i].flow); 62 if(v==vt){ 63 flow+=aug; 64 for(u=pre[v]; v!=vs; v=u,u=pre[u]){ 65 edge[cur[u]].flow+=aug; 66 edge[cur[u]^1].flow-=aug; 67 } 68 //aug=-1; 69 aug=INF; 70 } 71 break; 72 } 73 } 74 if(flag) continue; 75 int minlevel=NV; 76 for(int i=head[u]; i!=-1; i=edge[i].next){ 77 int v=edge[i].v; 78 if(edge[i].cap!=edge[i].flow && level[v]<minlevel){ 79 minlevel=level[v]; 80 cur[u]=i; 81 } 82 } 83 if(--gap[level[u]]==0) break; 84 level[u]=minlevel+1; 85 gap[level[u]]++; 86 u=pre[u]; 87 } 88 return flow; 89 } 90 91 int ans[111][111]; 92 int main(){ 93 int n,m; 94 scanf("%d%d",&n,&m); 95 vs=0; vt=2*n+1; NV=vt+1; NE=0; 96 memset(head,-1,sizeof(head)); 97 int a,b; 98 int tot1=0,tot2=0; 99 for(int i=1; i<=n; ++i){ 100 scanf("%d",&a); 101 addEdge(vs,i,a); 102 tot1+=a; 103 } 104 for(int i=1; i<=n; ++i){ 105 scanf("%d",&a); 106 addEdge(i+n,vt,a); 107 tot2+=a; 108 } 109 int tag=NE; 110 for(int i=1; i<=n; ++i){ 111 addEdge(i,i+n,INF); 112 } 113 while(m--){ 114 scanf("%d%d",&a,&b); 115 addEdge(a,b+n,INF); 116 addEdge(b,a+n,INF); 117 } 118 if(tot1==tot2 && ISAP()==tot2){ 119 puts("YES"); 120 for(int i=tag; i<NE; i+=2){ 121 int u=edge[i^1].v,v=edge[i].v-n; 122 ans[u][v]=edge[i].flow; 123 } 124 for(int i=1; i<=n; ++i){ 125 for(int j=1; j<=n; ++j){ 126 printf("%d ",ans[i][j]); 127 } 128 putchar(‘\n‘); 129 } 130 }else{ 131 puts("NO"); 132 } 133 return 0; 134 }
Codeforces 546E Soldier and Traveling(最大流)
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原文地址:http://www.cnblogs.com/WABoss/p/5668357.html