分析:最大权闭合图的裸题。
代码:
//Isap算法,复杂度O(n^2m)
#pragma comment(linker,"/STACK:102400000,102400000")
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll; //记得必要的时候改成无符号
const int maxn=505;
const int maxm=1000005;
const int INF=1000000000;
struct EdgeNode
{
int from;
int to;
int cost;
int next;
}edge[maxm];
int head[maxn],cnt;
void add(int x,int y,int z)
{
edge[cnt].from=x;edge[cnt].to=y;edge[cnt].cost=z;edge[cnt].next=head[x];head[x]=cnt++;
edge[cnt].from=y;edge[cnt].to=x;edge[cnt].cost=0;edge[cnt].next=head[y];head[y]=cnt++;
}
void init()
{
cnt=0;
memset(head,-1,sizeof(head));
}
int S,T,n,m;
int d[maxn],gap[maxn],curedge[maxn],pre[maxn];
//curedge[]为当前弧数组,pre为前驱数组
int sap(int S,int T,int n) //n为点数
{
int cur_flow,flow_ans=0,u,tmp,neck,i;
memset(d,0,sizeof(d));
memset(gap,0,sizeof(gap));
memset(pre,-1,sizeof(pre));
for(i=0;i<=n;i++)curedge[i]=head[i]; //初始化当前弧为第一条邻接表
gap[0]=n;
u=S;
while(d[S]<n) //当d[S]>=n时,网络中肯定出现了断层
{
if(u==T)
{
cur_flow=INF;
for(i=S;i!=T;i=edge[curedge[i]].to)
{ //增广成功,寻找瓶颈边
if(cur_flow>edge[curedge[i]].cost)
{
neck=i;
cur_flow=edge[curedge[i]].cost;
}
}
for(i=S;i!=T;i=edge[curedge[i]].to)
{ //修改路径上的边容量
tmp=curedge[i];
edge[tmp].cost-=cur_flow;
edge[tmp^1].cost+=cur_flow;
}
flow_ans+=cur_flow;
u=neck; //下次增广从瓶颈边开始
}
for(i=curedge[u];i!=-1;i=edge[i].next)
if(edge[i].cost&&d[u]==d[edge[i].to]+1)
break;
if(i!=-1)
{
curedge[u]=i;
pre[edge[i].to]=u;
u=edge[i].to;
}
else
{
if(0==--gap[d[u]])break; //gap优化
curedge[u]=head[u];
for(tmp=n,i=head[u];i!=-1;i=edge[i].next)
if(edge[i].cost)
tmp=min(tmp,d[edge[i].to]);
d[u]=tmp+1;
++gap[d[u]];
if(u!=S)u=pre[u]; //重标号并且从当前点前驱重新增广
}
}
return flow_ans;
}
int a[maxn];
int main()
{
int i,x,y,sum;
while(~scanf("%d%d",&n,&m))
{
init(); S=0; T=n+1; sum=0;
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
if(a[i]>=0){
sum+=a[i];
add(S,i,a[i]);
}
else add(i,T,-a[i]);
}
for(i=1;i<=m;i++){
scanf("%d%d",&x,&y);
add(x,y,INF);
}
n=T+1;
printf("%d\n",sum-sap(S,T,n));
}
return 0;
}