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;
}