CountHunter 6101 最优贸易 强联通缩点

题目传送门

题解：强连通锁点之后。 就成了一副单向图。 然后对于每个点 找到 后面合法的点的最大值就好了。 合法就是后面的那个点可以走到n号点。

也可以正向跑一遍dij 求出到这个点的最小花费。 然后在反向跑dij跑出n到这个点的最大花费，然后枚举每个点。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+7;
const int N = 1e5 + 100;
const int M = 1e6 + 100;
int head[N], to[M], nt[M], tot = 0;
int a[N];
vector<int> vc[N];
void add(int u, int v){
    to[tot] = v;
    nt[tot] = head[u];
    head[u] = tot++;
}
int n, m, u, v, op;
int belong[N], dfn[N], low[N], now_time, scc_cnt;
int mx[N], mn[N], ret[N];
stack<int> s;
void dfs(int u){
    dfn[u] = low[u] = ++now_time;
    s.push(u);
    for(int i = head[u]; ~i; i = nt[i]){
        if(!dfn[to[i]]) dfs(to[i]);
        if(!belong[to[i]]) low[u] = min(low[u], low[to[i]]);
    }
    if(dfn[u] == low[u]){
        ++scc_cnt;
        int now;
        while(1){
            now = s.top(); s.pop();
            belong[now] = scc_cnt;
            if(now == u) break;
        }
    }
}
void scc(int n){
    now_time = scc_cnt = 0;
    for(int i = 1; i <= n; ++i)
        if(!belong[i]) dfs(i);
    int v;
    for(int i = 1; i <= n; ++i){
        for(int j = head[i]; ~j; j=nt[j]){
            v = to[j];
            if(belong[v] != belong[i]){
                    vc[belong[i]].pb(belong[v]);
            }
        }
    }
    for(int i = 1; i <= n; ++i){
        int rt = belong[i];
        if(ret[rt] == 0){
            ret[rt] = -1;
            mn[rt] = a[i];
            mx[rt] = a[i];
        }
        else {
            mn[rt] = min(mn[rt], a[i]);
            mx[rt] = max(mx[rt], a[i]);
        }
    }
}
int ans = 0;
int solve(int x){
    if(ret[x] != -1) return mx[x];
    ret[x] = 0;
    int mmax = -1;
    if(x == belong[n]) mmax = mx[x];
    for(int v : vc[x]){
        mmax = max(solve(v), mmax);
    }
    if(mmax != -1) {
        mx[x] = max(mx[x], mmax);
        ans = max(ans, mx[x]-mn[x]);
    }
    else mx[x] = -1;
    return mx[x];
}
int main(){
    memset(head, -1, sizeof(head));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for(int i = 1; i <= m; ++i){
        scanf("%d%d%d", &u, &v, &op);
        add(u, v);
        if(op == 2) add(v, u);
    }
    scc(n);
    solve(belong[1]);
    cout << ans << endl;
    return 0;
}