uestc 方老师抢银行

参考的大牛的思路和代码

思路： 每次经过一个强连通分量时，都可以走完该分量从而使抢得的钱最多，可以把每个强连通分量缩点，用总的钱数代表该点的值。然后重新遍历新图，遇到网吧可以更新一下结果。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#include<climits>
#include<string.h>
#include<cmath>
#include<stdlib.h>
#include<vector>
#include<stack>
#include<set>
using namespace std;
#define INF 1000000007
#define MAXN 300010
#define Mod 1000007
#define N 100007
#define NN 30
#define sigma_size 3
const int maxn = 6e5 + 10;
using namespace std;
typedef long long LL;
const double pi = acos(-1);

vector<int> G[MAXN], G2[MAXN];
int dfn[N], lowlink[N], instk[N], bel[N], coin[N], iswb[N], wb[N], coin2[N];
int n, m, dfs_clock, scc_cnt, p, k, res;
int u, v;
stack<int> S;

void Tarjan(int u)
{
    dfn[u] = lowlink[u] = ++dfs_clock;
    S.push(u);
    instk[u] = 1;
    for (int i = 0; i < G[u].size(); ++i) {
        int v = G[u][i];
        if (!dfn[v]) {
            Tarjan(v);
            lowlink[u] = min(lowlink[u], lowlink[v]);
        }
        else if (instk[v]) {
            lowlink[u] = min(lowlink[u], dfn[v]);
        }
    }
    if (lowlink[u] == dfn[u]) {            //找到一个强连通分量
        scc_cnt++;                        //强连通分量数目加一
        int v;
        do {
            v = S.top();
            S.pop();
            if (iswb[v]) wb[scc_cnt] = 1;
            instk[v] = 0;
            coin2[scc_cnt] += coin[v];
            bel[v] = scc_cnt;
            if (v == p) p = scc_cnt;
        } while (v != u);
    }
}

void find_scc()                            //Tarjan求强连通分量
{
    for (int i = 1; i <= n; ++i)
        if (!dfn[i]) Tarjan(i);
}

void dfs(int u, int sum)                //遍历新建的图
{
    sum += coin2[u];
    if (wb[u]) res = sum;
    for (int i = 0; i < G2[u].size(); ++i)
        dfs(G2[u][i], sum);
}

void init()
{
    for (int i = 0; i <= n; ++i) {
        G[i].clear();
        G2[i].clear();
    }
    while (!S.empty()) S.pop();
    dfs_clock = scc_cnt = 0;
    memset(instk, 0, sizeof(instk));
    memset(dfn, 0, sizeof(dfn));
    memset(iswb, 0, sizeof(iswb));
    memset(wb, 0, sizeof(wb));
    memset(bel, -1, sizeof(bel));
    memset(coin2, 0, sizeof(coin2));
    memset(lowlink, 0, sizeof(lowlink));
}

void read()
{
    for (int i = 0; i < m; ++i) {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
    }
    for (int i = 1; i <= n; ++i)
        scanf("%d", &coin[i]);
    scanf("%d%d", &p, &k);
    for (int i = 0; i < k; ++i) {
        scanf("%d", &v);
        iswb[v] = 1;
    }
}

void rebuild()                                //根据强连通分量重新建图
{
    for (int i = 1; i <= n; ++i) {
        for (int j = 0; j < G[i].size(); ++j) {
            u = bel[i];
            v = bel[G[i][j]];
            if (u != v)
                G2[u].push_back(v);
        }
    }
    res = -1;
    dfs(p, 0);
    cout << res << endl;
}

void process()
{
    init();
    read();
    find_scc();
    rebuild();
}

int main()
{
    while (cin >> n >> m)
        process();
    //system("pause");
    return 0;
}