【BZOJ 1179】[Apio2009]Atm

【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

tarjan强连通缩点一下。
然后把缩点之后,每个点的钱的数累加起来。
然后从S出发
开始一边做bfs一遍做dp.
最后输出有酒吧的点的dp值中的最大值。

【代码】

/*
    n个点,m条有向边.
    把有向图G的环进行缩点;
    缩完之后的图存在vector <int> g[N]里面;
    n变为缩完点之后的图的节点的个数了。
*/
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 5e5;//节点个数

vector <int> G[N+10],g[N+10];
int n,m,nn,tot = 0,top = 0,dfn[N+10],low[N+10],z[N+10],totn,in[N+10],value[N+10],value1[N+10];
int s,p,bo[N+10],bo1[N+10],dp[N+10];
int bh[N+10];
queue<int> dl;

void dfs(int x){
    dfn[x] = low[x] = ++ tot;
    z[++top] = x;
    in[x] = 1;
    int len = G[x].size();
    rep1(i,0,len-1){
        int y = G[x][i];
        if (!dfn[y]){
            dfs(y);
            low[x] = min(low[x],low[y]);
        }else
        if (in[y] && dfn[y]<low[x]){
            low[x] = dfn[y];
        }
    }
    if (low[x]==dfn[x]){
        int v = 0;
        totn++;
        while (v!=x){
            v = z[top];
            in[v] = 0;
            bh[v] = totn;
            top--;
        }
    }
}

int main(){
    #ifdef LOCAL_DEFINE
        freopen("rush_in.txt", "r", stdin);
    #endif

    tot = 0,totn = 0;
    ri(n),ri(m);

    rep1(i,1,m){
        int x,y;
        ri(x),ri(y);
        G[x].pb(y);
    }

    rep1(i,1,n) ri(value[i]);

    ri(s),ri(p);
    rep1(i,1,p){
        int x;
        ri(x);
        bo[x] = 1;
    }

    rep1(i,1,n)
        if (dfn[i]==0)
            dfs(i);

    nn = totn;

    rep1(i,1,n){
        int len = G[i].size();
        int xx = bh[i];
        rep1(j,0,len-1){
            int y = G[i][j];
            int yy = bh[y];
            if (xx!=yy)
                g[xx].pb(yy);
        }
    }

    rep1(i,1,n){
        if (bo[i])
            bo1[bh[i]] = 1;
        value1[bh[i]]+=value[i];
    }

    memset(dp,255,sizeof dp);
    dl.push(bh[s]);
    dp[bh[s]] = value1[bh[s]];
    while (!dl.empty()){
        int x = dl.front();
        dl.pop();
        int len = g[x].size();
        rep1(i,0,len-1){
            int y = g[x][i];
            if (dp[y]<dp[x]+value1[y]){
                dp[y] = dp[x] + value1[y];
                dl.push(y);
            }
        }
    }

    int ans = 0;
    rep1(i,1,nn)
    if (bo1[i]){
        ans = max(ans,dp[i]);
    }
    printf("%d\n",ans);
    return 0;
}

原文地址：https://www.cnblogs.com/AWCXV/p/8669890.html