洛谷 1262 间谍网络 Tarjan 图论

#洛谷 1262 图论 tarjan

并不感觉把这道题目放在图的遍历中很合适，虽然思路比较简单但是代码还是有点多的，，

将可收买的间谍的cost值设为它的价格，不可购买的设为inf，按照控制关系连图，Tarjan缩点，得到的新图中，入度为0的点是必须购买的，如果这些点中存在inf，则不成立

#include <cstdio>
#include <algorithm>
#include <cstring>

const int inf = 0x3f3f3f3f;
const int maxn = (3000 + 500) * 2;
const int maxm =  50000;
int dfn[maxn], low[maxn], vis[maxn];
int sta[maxn];
int stackTop = 0;
int tim = 0;
int cost[maxn];
int costNew[maxn];
int from[maxn];
int cur = 0;
int indo[maxn];
int outdo[maxn];
int n, p, r;
int x, y;
int last[maxm], other[maxm], pre[maxm];
int totEdge = 0;

void add(int x, int y) {
    totEdge++;
    pre[totEdge] = last[x];
    last[x] = totEdge;
    other[totEdge] = y;
}
void tarjan(int x) {
    tim++;
    dfn[x] = low[x] = tim;
    vis[x] = 1;
    stackTop++;
    sta[stackTop] = x;
    for (int p = last[x]; p; p = pre[p]) {
        int q = other[p];
        if (!dfn[q]) {
            tarjan(q);
            low[x] = std :: min(low[x], low[q]);
        } else if (vis[q]) {
            low[x] = std :: min(low[x], dfn[q]);
        }
    }
    if (dfn[x] == low[x]) {
        cur++;
        costNew[cur] = inf;
        while (sta[stackTop] != x) {
            vis[sta[stackTop]] = 0;
            from[sta[stackTop]] = cur;
            costNew[cur] = std :: min(costNew[cur], cost[sta[stackTop]]);
            stackTop--;
        }
        vis[x] = 0;
        from[x] = cur;
        costNew[cur] = std :: min(costNew[cur], cost[x]);
        stackTop--;
    }
}

int main () {
    scanf("%d", &n);
    scanf("%d", &p);
    for (int i = 1; i <= p; i++) {
        scanf("%d", &x);
        scanf("%d", &cost[x]);
    }
    for (int i = 1; i <= n; i++)
        if (cost[i] == 0) cost[i] = inf;
    scanf("%d", &r);
    while (r--) {
        scanf("%d %d", &x, &y);
        add(x, y);
    }
    for (int i = 1; i <= n; i++) {
        if (!dfn[i]) tarjan(i);
    }
    for (int i = 1; i <= n; i++)
        for (int j = last[i]; j; j = pre[j]) {
            int q = other[j];
            if (from[i] != from[q]) {
                indo[from[q]]++;
                outdo[from[i]]++;
            }
        }
    long long ans = 0;
    bool flag = 0;
    for (int i = 1; i <= cur; i++) {
        if (indo[i] == 0) {
            if (costNew[i] == inf) {
                flag = 1;
                break;
            } else {
                ans += costNew[i];
            }
        }
    }
    if (!flag) {
        printf("YES\n%lld\n", ans);
    } else {
        printf("NO\n");
        for (int i = 1; i <= n; i++) {
            if (indo[from[i]] == 0 && costNew[from[i]] == inf) {
                printf("%d", i);
                break;
            }
        }
    }
    return 0;
}