1179: [Apio2009]Atm

Description

Input

第一行包含两个整数N、M。N表示路口的个数，M表示道路条数。接下来M行，每行两个整数，这两个整数都在1到N之间，第i+1行的两个整数表示第i条道路的起点和终点的路口编号。接下来N行，每行一个整数，按顺序表示每个路口处的ATM机中的钱数。接下来一行包含两个整数S、P，S表示市中心的编号，也就是出发的路口。P表示酒吧数目。接下来的一行中有P个整数，表示P个有酒吧的路口的编号

Output

输出一个整数，表示Banditji从市中心开始到某个酒吧结束所能抢劫的最多的现金总数。

Sample Input

6 7
1 2
2 3
3 5
2 4
4 1
2 6
6 5
10
12
8
16
1 5
1 4
4
3
5
6

Sample Output

47

求出强连通分量缩成点形成一个新图，将所有带酒吧的点连到一个重点，然后spfa

#include<stdio.h>#include<algorithm>#include<string.h>#include<vector>#include<stack>#include<queue>using namespace std;#define ll long longconst int maxm = 500005;stack<int>s;vector<int>v[maxm];vector<int>p[maxm];int id = 1, sum = 1, low[maxm], dfn[maxm], b[maxm];int num[maxm], vis[maxm], flag1[maxm], a[maxm], flag2[maxm];ll ans = 0, f[maxm], dis[maxm];typedef struct{int x, y;}H;H r[maxm];void tarjian(int x);void bfs(int s);int main(){int n, i, j, k, m, t, s;scanf("%d%d", &n, &m);for (i = 1;i <= m;i++){ scanf("%d%d", &r[i].x, &r[i].y);v[r[i].x].push_back(r[i].y);}for (i = 1;i <= n;i++)scanf("%d", &a[i]);scanf("%d%d", &s, &t);for (i = 1;i <= t;i++){scanf("%d", &k);flag1[k] = 1;}sum = 1, ans = 0;for (i = 1;i <= n;i++){if (!dfn[i])tarjian(i);}for (i = 1;i <= m;i++){if (b[r[i].x] != b[r[i].y])p[b[r[i].x]].push_back(b[r[i].y]);}for (i = 1;i < sum;i++){if (flag2[i])p[i].push_back(0);}memset(vis, 0, sizeof(vis));bfs(b[s]);printf("%d\n", dis[0]);return 0;}void tarjian(int x){int i, j;vis[x] = 1, low[x] = dfn[x] = id++;s.push(x);for (i = 0;i < v[x].size();i++){int xx = v[x][i];if (!dfn[xx]){tarjian(xx);low[x] = min(low[x], low[xx]);}else if (vis[xx])low[x] = min(low[x], dfn[xx]);}if (low[x] == dfn[x]){while (s.size()){int xx = s.top();s.pop();b[xx] = sum;if (flag1[xx])flag2[sum] = 1;f[sum] += a[xx];vis[xx] = 0;if (xx == x)break;}sum++;}}void bfs(int s){queue<int>q;memset(dis, 0, sizeof(dis));dis[s] = f[s];q.push(s);while (!q.empty()){int xx = q.front();q.pop();for (int i = 0;i < p[xx].size();i++){if (dis[p[xx][i]] < dis[xx] + f[p[xx][i]]){dis[p[xx][i]] = dis[xx] + f[p[xx][i]];q.push(p[xx][i]);}}}}