题目大意：给出一个张有向图，和N个点的价值（买了这个点后，以这个点出发的所能遍及的点都会被染色）
问至少要花费多少钱去买点，才能使得这张图的所有点都被染色 如果不能所有的点都染色，输出不能染色点的最小值
解题思路：先求出所有的强连通分量，接着求出每个强连通分量内所有点的最少价值，因为同一个强连通分量内的点只需要购买一个就可以全部染色了 接着缩点，用桥连接起来，，形成一张新的图(以下所说的点都是强连通分量的缩点) 找出所有入度为0的点，因为入度为0的点是无法通过别的点进行染色的，所以只能买入度为0的点
N 3010min(a,b) ((a) < (b) ? (a) : (b))struct Edge{int from, to, next;}E[M];int n, tot, p, m, dfs_clock, scc_cnt, top;int head[N], cost[N], pre[N], sccno[N], stack[N], lowlink[N], in[N], value[N];void AddEdge(int u, int v) {E[tot].to = v;E[tot].from = u;E[tot].next = head[u];head[u] = tot++;}void init() {memset(cost, 0x3f, sizeof(cost));scanf(“%d”, &p);int No, money;for (int i = 0; i < p; i++) {scanf(“%d%d”, &No, &money);cost[No] = money;}scanf(“%d”, &m);memset(head, -1, sizeof(head));tot = 0;int u, v;for (int i = 0; i < m; i++) {scanf(“%d%d”, &u, &v);AddEdge(u, v);}}void dfs(int u) {pre[u] = lowlink[u] = ++dfs_clock;stack[++top] = u;int v;for (int i = head[u]; i != -1; i = E[i].next) {v = E[i].to;if (!pre[v]) {dfs(v);lowlink[u] = min(lowlink[u], lowlink[v]);}else if (!sccno[v]) {lowlink[u] = min(lowlink[u], pre[v]);}}if (lowlink[u] == pre[u]) {scc_cnt++;while (1) {v = stack[top–];sccno[v] = scc_cnt;if (v == u)break;}}}void solve() {memset(pre, 0, sizeof(pre));memset(sccno, 0, sizeof(sccno));dfs_clock = scc_cnt = top = 0;for (int i = 1; i <= n; i++)if (!pre[i])dfs(i);for (int i = 1; i <= scc_cnt; i++) {value[i] = INF;in[i] = 1;}for (int i = 1; i <= n; i++) {value[sccno[i]] = min(value[sccno[i]], cost[i]);}int u, v, minid;for (int i = 0; i < tot; i++) {u = sccno[E[i].from];v = sccno[E[i].to];if (u != v) {in[v] = 0;}}bool flag = false;for (int i = 1; i <= n; i++) {u = sccno[i];if (in[u] && value[u] == INF) {minid = i;flag = true;break;}}if (flag) {printf(“NO\n%d\n”, minid);return ;}int ans = 0;for (int i = 1; i <= scc_cnt; i++) {if (in[i])ans += value[i];}printf(“YES\n%d\n”, ans);}int main() {while (scanf(“%d”, &n) != EOF) {init();solve();}return 0;}
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