[破碎的状态] [-9] 546E
网络流模版题
曾经在去年救过我省选的题,感谢周植
题意:有n个城市,第i个城市里一开始有ai个人
后来,有些人往某个相邻的城市走(只走一次),之后,第i个城市有bi个人
求是否有可能,如果可能,输出方案数,也就是说第i个城市有几个人往第j个城市走(i=j则表示没走的人)
这真的是网络流..不是最小割..
第i个城市向第j个连边,然后流一遍..求最大流..
代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 | #include<set>#include<map>#include<list>#include<queue>#include<stack>#include<string>#include<time.h>#include<math.h>#include<memory>#include<vector>#include<bitset>#include<fstream>#include<stdio.h>#include<utility>#include<sstream>#include<string.h>#include<iostream>#include<stdlib.h>#include<algorithm>using namespace std;int a[105],b[105];int ans[105][105];const int maxn=505;const int inf=99999999;struct edge{ int y; int weight; edge * next; edge * anti;};edge * li[1005];edge * new_edge(){ static edge a[100005]; static int top=0; return &a[top++];}edge * inserts(int x,int y,int z){ edge * t=new_edge(); t->y=y; t->weight=z; t->next=li[x]; li[x]=t; return t;}void insert_edge(int x,int y,int z){ edge * t1=inserts(x,y,z); edge * t2=inserts(y,x,0); t1->anti=t2; t2->anti=t1;}int dis[maxn];bool bfs(int s,int t){ static int que[maxn]; int front=0,rail=1; memset(dis,-1,sizeof(dis)); dis[s]=0; que[0]=s; for (;front<rail;front++) { int now=que[front]; edge * x; for (x=li[now];x!=0;x=x->next) { if ((dis[x->y]==-1)&&(x->weight!=0)) { dis[x->y]=dis[now]+1; if (x->y==t) return true; que[rail++]=x->y; } } } return false;}int max_flow(int s,int t){ int tot=0; for (;bfs(s,t);) { static int pre[maxn]; static edge * cur[maxn]; static edge * path[maxn]; memcpy(cur,li,sizeof(cur)); int now=s; for (;dis[s]!=-1;) { if (now==t) { int temp; int mini=inf; for (temp=t;temp!=s;temp=pre[temp]) { mini=min(mini,path[temp]->weight); } tot+=mini; for (temp=t;temp!=s;temp=pre[temp]) { path[temp]->weight-=mini; path[temp]->anti->weight+=mini; } now=s; } edge * x; for (x=li[now];x!=0;x=x->next) { if ((x->weight!=0)&&(dis[x->y]==dis[now]+1)) { pre[x->y]=now; path[x->y]=x; cur[now]=x; now=x->y; break; } } if (x==0) { dis[now]=-1; now=pre[now]; } } } return tot;}int main(){ #ifdef absi2011 freopen("input.txt","r",stdin); freopen("output.txt","w",stdout); #endif int n,m; scanf("%d%d",&n,&m); int i; int sum_a=0; for (i=0;i<n;i++) { scanf("%d",&a[i]); sum_a+=a[i]; } int sum_b=0; for (i=0;i<n;i++) { scanf("%d",&b[i]); sum_b+=b[i]; } if (sum_a!=sum_b) { printf("NO\n"); return 0; } int s=n*2,t=n*2+1; for (i=0;i<n;i++) { insert_edge(s,i,a[i]); } for (i=0;i<n;i++) { insert_edge(i+n,t,b[i]); insert_edge(i,i+n,inf); } for (i=0;i<m;i++) { int x,y; scanf("%d%d",&x,&y); x--; y--; insert_edge(x,y+n,inf); insert_edge(y,x+n,inf); } int val=max_flow(s,t); if (val!=sum_a) { printf("NO\n"); } else { printf("YES\n"); for (i=0;i<n;i++) { edge * x; for (x=li[i+n];x!=0;x=x->next) { if (x->y==t) continue; ans[x->y][i]+=x->weight; } } for (i=0;i<n;i++) { int j; for (j=0;j<n;j++) { printf("%d ",ans[i][j]); } printf("\n"); } return 0; } return 0;} |
评论 (0)