【35.00%】【z13】&&【b093】最优贸易

【题解】
这题就是要在n个点里面选一个花费最小的点。然后找一个花费最大的点。两者之差为最大值。
但是最大值的点要在最小值的点之后出现。且走到后者之后要能够到达N号节点。为了处理掉环。先用tarjan进行缩点。这样整张图就不会出现环了。接下来进行DP即可。记录到达某个点之前所能买到的最低价格。然后尝试在这个点卖掉。记录最优值。然后把最优值一直往后传。最后输出f[n]。这里的DP用了记忆化搜索。或者说是一个强力剪枝。自己看吧。
【代码】

//#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <vector>
#include <algorithm>
#include <stack>

using namespace std;

const int MAXN = 100000 + 100;

struct data2
{
    int sell, min;
};

int n, m, dx = 0, nn = 0, bh[MAXN];
int price[MAXN], dfn[MAXN], low[MAXN], ma_x[MAXN], mi_n[MAXN], ans = 0;
data2 f[MAXN];
vector <int> a[MAXN], aa[MAXN];
stack <int> z;
int zzz[MAXN];
bool flag[MAXN] = { 0 };
bool used[MAXN] = { 0 };

void tarjan(int x)
{
    flag[x] = true;
    dfn[x] = low[x] = ++dx;
    zzz[x] = z.size();
    z.push(x);
    int len = a[x].size();
    for (int i = 0; i <= len - 1; i++)
    {
        int y = a[x][i];
        if (!flag[y])
        {
            tarjan(y);
            low[x] = min(low[x], low[y]);
        }
        else
            if (zzz[y] >0)
                low[x] = min(low[x], dfn[y]);
    }
    if (low[x] == dfn[x])
    {
        nn++;
        int tma = price[x], tmi = price[x], limit = zzz[x];
        while (z.size() != limit)
        {
            int t = z.top();
            tma = max(tma, price[t]);
            tmi = min(tmi, price[t]);
            bh[t] = nn;
            zzz[t] = 0;
            z.pop();
        }
        ma_x[nn] = tma;
        mi_n[nn] = tmi;
    }
}

void dfs(int now, int pre)
{
    bool can = false; //这个can表示这个点有没有更新。如果没有更新任何东西就剪枝
    if (f[now].min > f[pre].min)
    {
        f[now].min = f[pre].min;
        can = true;
    }
    if (f[now].min > mi_n[now])
    {
        f[now].min = mi_n[now];
        can = true;
    }
    if (f[now].sell < ma_x[now] - f[pre].min)
    {
        f[now].sell = ma_x[now] - f[pre].min;
        can = true;
    }
    if (f[now].sell < ma_x[now] - mi_n[now])
    {
        f[now].sell = ma_x[now] - mi_n[now];
        can = true;
    }
    if (f[now].sell < f[pre].sell)//答案往后传
    {
        f[now].sell = f[pre].sell;
        can = true;
    }
    if (!can)
        return;
    int len = aa[now].size();
    for (int i = 0; i <= len - 1; i++)
        if (!used[aa[now][i]])
        {
            used[aa[now][i]] = true;
            dfs(aa[now][i], now);
            used[aa[now][i]] = false;
        }
}

int main()
{
    //freopen("F:\rush.txt", "r", stdin);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++)
        scanf("%d", &price[i]);
    for (int i = 1; i <= m; i++)
    {
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        a[x].push_back(y);
        if (z == 2) a[y].push_back(x);
    }
    tarjan(1);
    for (int i = 1; i <= n; i++)
        if (bh[i] != 0)//bh数组记录了缩点后i号点新的编号
        {
            int len = a[i].size();
            for (int j = 0; j <= len - 1; j++)
            {
                int y = a[i][j];
                if (bh[y] != 0 && bh[i] != bh[y])
                    aa[bh[i]].push_back(bh[y]);
            }
        }
    for (int i = 1; i <= nn; i++)
        f[i].min = 2100000000;
    int qidian = bh[1];//从起点开始dfs+dp
    used[qidian] = true;
    f[qidian].min = mi_n[qidian];//一开始就可能在环里面
    f[qidian].sell = ma_x[qidian] - mi_n[qidian];
    int len = aa[qidian].size();
    for (int i = 0; i <= len - 1; i++)
        if (!used[aa[qidian][i]])
        {
            used[aa[qidian][i]] = true;
            dfs(aa[qidian][i], qidian);
            used[aa[qidian][i]] = false;
        }
    printf("%d
", f[bh[n]].sell);
    return 0;
}