UVALive 6206

这是个图论问题

做法是首先缩点，环内的边的删除不会影响连通性，缩点得到一棵树，然后dfs计算树边的新权值，接下来二分搜索最大话费，在dfs贪心判断是否可行即可。总之这道题是个大综合题。。。

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <map>#include <stack>#include <vector>#define N 10100#define M 20100typedef long long ll;using namespace std;struct Edge{    int v,next;    ll w;}edge[M*2];int head[N],n,m,cnt;ll val[N],mid;int scc,depth,dfn[N],low[N],belong[N],tot[N];bool vis[N],flag;stack<int>sta;vector<int>ve[N];void init(){    memset(head,-1,sizeof(head));    memset(dfn,0,sizeof(dfn));    memset(tot,0,sizeof(tot));    cnt=depth=scc=0;    for(int i=1;i<=n;i++) ve[i].clear();}void addedge(int u,int v,ll w){    edge[cnt].v=v;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].v=u;    edge[cnt].w=w;    edge[cnt].next=head[v];    head[v]=cnt++;}void tarjan(int u,int fa){    dfn[u]=low[u]=++depth;    sta.push(u);    for(int j=head[u];j!=-1;j=edge[j].next){        int v=edge[j].v;        if(v==fa) continue;        if(dfn[v]==0){            tarjan(v,u);            low[u]=min(low[v],low[u]);        }        else low[u]=min(low[u],dfn[v]);    }    if(dfn[u]==low[u]){        scc++;        while(1){            int tem=sta.top();            sta.pop();            belong[tem]=scc;            ve[scc].push_back(tem);            if(tem==u)break;        }    }}int dfs(int u,int total){   // 有vis数组不需要fa变量    int i,j,sum,tem;    ll w;    vis[belong[u]]=1;    sum=ve[belong[u]].size();    for(i=0;i<ve[belong[u]].size();i++){        int v=ve[belong[u]][i];        for(j=head[v];j!=-1;j=edge[j].next){            int vv=edge[j].v;            if(vis[belong[vv]])continue;            if(belong[vv]==belong[v])continue;            tem=dfs(vv,total);            sum+=tem;            w=edge[j].w;            edge[j].w=edge[j^1].w=w*tem*(total-tem);        }    }    return sum;}bool dfs_check(int u,ll dis){ // 有vis数组不需要fa变量    int i,j;    ll sum;    bool ok;    vis[belong[u]]=1;    for(i=0;i<ve[belong[u]].size();i++){        int v=ve[belong[u]][i];        sum=val[v];        for(j=head[v];j!=-1;j=edge[j].next){            int vv=edge[j].v;            if(vis[belong[vv]])continue;            if(belong[vv]==belong[v])continue;            if(dfs_check(vv,edge[j].w)==0)                sum+=edge[j].w;        }        if(sum>mid) flag=0;        if(v==u) {if(dis+sum>mid) ok=0; else ok=1;}    }    return ok;}bool check(){    memset(vis,0,sizeof(vis));    flag=1;    for(int i=1;i<=n;i++){        if(vis[belong[i]])continue;        dfs_check(i,0);        if(flag==0) return 0;    }    return 1;}int main(){    int t,T;    int u,v;    int i,j;    ll w;    scanf("%d",&T);    for(t=1;t<=T;t++){        scanf("%d %d",&n,&m);        for(i=1;i<=n;i++) scanf("%lld",&val[i]);        init();        for(i=1;i<=m;i++){            scanf("%d %d %lld",&u,&v,&w);            addedge(u,v,w);        }        for(i=1;i<=n;i++){            if(!dfn[i]){                depth=0;                tarjan(i,0);                tot[i]=depth;            }        }        memset(vis,0,sizeof(vis));        for(i=1;i<=n;i++){            if(vis[belong[i]])continue;            dfs(i,tot[i]);        }        ll l=0,r=1000000000000LL,ans;        while(l<=r){            mid=(l+r)>>1;            if(check()){                ans=mid;                r=mid-1;            }            else l=mid+1;        }        printf("Case %d: %lld\n",t,ans);    }    return 0;}