【算法练习】HihoCoder 1185 连通性·三 （强连通分量）

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/pengwill97/article/details/81951621

题意

从1为起点，找到一条有向路径，其经过节点和最大。

题解

tarjan算法缩点，形成有DAG，在DAG上DP即可，

代码

#include<bits/stdc++.h>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
int n, m;
int grass[nmax], ans[nmax], ss[nmax], nowans, st;
int head[nmax],head2[nmax], tot;
int dfn[nmax], low[nmax], color[nmax], scc, dfs_clock, tot2;
bool visit[nmax], instack[nmax], visitscc[nmax];
int newgrass[nmax];
struct edge {
    int to, nxt;
}e[nmax << 1], e2[nmax << 1];
void add_edge(int u, int v) {
    e[tot].to = v;
    e[tot].nxt = head[u];
    head[u] = tot ++;
}
void add_edge2(int u, int v) {
    e2[tot2].to = v;
    e2[tot2].nxt = head2[u];
    head2[u] = tot2 ++;
}
void tarjan(int u) {
    dfn[u] = low[u] = ++dfs_clock;
    ss[st++] = u;
    instack[u] = true;
    for(int i = head[u]; i != -1; i = e[i].nxt) {
        int v = e[i].to;
        if(!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        } else if(instack[v]) {
            low[u] = min(low[u], dfn[v]);
        }
    }

    if(low[u] == dfn[u]) {
        int temp = 0, now = 0;
        scc ++;
        while(now != u){
            now = ss[--st];
            instack[now] = false;
            color[now] = scc;
            temp +=  grass[now];
        }
        newgrass[scc] = temp;
    }
}
void dfs(int u) {
    visit[u] = true;
    for(int i = head[u]; i != -1; i = e[i].nxt) {
        int v = e[i].to;
        if(!visit[v]) {
            if(color[u] != color[v] ){
                add_edge2(color[u], color[v]);
                dfs(v);
            } else
                dfs(v);
        } else if(color[u] != color[v]) {
            add_edge2(color[u], color[v]);
        }
    }
}

void dfs2(int u) {
    if(head2[u] != -1) {
        for(int i = head2[u]; i != -1; i = e2[i].nxt) {
            int v = e2[i].to;
            dfs2(v);
            ans[u] = max(ans[u], ans[v] + newgrass[u]);
        }
    } else {
        ans[u] = newgrass[u];
    }

}

int main(){
//    freopen("in.txt", "r", stdin);
    memset(head, -1, sizeof head);
    memset(head2, -1, sizeof head2);
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; ++i) {
        scanf("%d", &grass[i]);
    }
    for(int i = 1; i <= m; ++i) {
        int u, v;
        scanf("%d %d", &u, &v);
        add_edge(u, v);
    }
    tarjan(1);
    dfs(1);
    dfs2(color[1]);
    printf("%d\n", ans[color[1]]);

    return 0;
}