Tarjan缩点
P3387 【模板】缩点
思路
既然时缩点的模板，那么缩点自然少不了了，缩点后我们的到新的有向无环图，然后再利用这个无环图去找一条最大权值的路径，路径和即为答案。
我们改如何选取起点来避免不必要的计算，假设存在一条路径，我们的最大值一定时从起点开始的，所以我们选取所有的缩点以后入度为零的点去bfs，然后不断更新最大路径值。
代码
#include
using namespace std;
typedef long long ll;
inline ll read() {
ll f = 1, x = 0;
char c = getchar();
while(c > '9' || c < '0') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f * x;
}
const int N1 = 1e4 + 10, N2 = 1e5 + 10;
int head[N1], to[N2], nex[N2], cnt = 1;
int dfn[N1], low[N1], visit[N1], scc[N1], point[N1], value[N1], n, m, tot, sum;
int x[N2], y[N2];
int stk[N1], top;
int head1[N1], to1[N2], nex1[N2], cnt1 = 1;
int in[N1];
void tarjan(int rt) {
dfn[rt] = low[rt] = ++tot;
visit[rt] = 1;
stk[++top] = rt;
for(int i = head[rt]; i; i = nex[i]) {
if(!dfn[to[i]]) {
tarjan(to[i]);
low[rt] = min(low[rt], low[to[i]]);
}
else if(visit[to[i]]) {
low[rt] = min(low[rt], dfn[to[i]]);
}
}
if(dfn[rt] == low[rt]) {
sum++;
do {
visit[stk[top]] = 0;
scc[stk[top]] = sum;
value[sum] += point[stk[top]];
top--;
}while(stk[top + 1] != rt);
}
}
void add(int x, int y) {
to[cnt] = y;
nex[cnt] = head[x];
head[x] = cnt++;
}
void bfs() {
queue> q;
for(int i = 1; i <= sum; i++)
if(in[i] == 0)
q.push(make_pair(i, value[i]));
int ans = 0;
while(!q.empty()) {
int temp = q.front().first;
ans = max(ans, q.front().second);
for(int i = head1[temp]; i; i = nex1[i])
q.push(make_pair(to1[i], q.front().second + value[to1[i]]));
q.pop();
}
printf("%d\n", ans);
}
int main() {
// freopen("in.txt", "r", stdin);
n = read(), m = read();
for(int i = 1; i <= n; i++)
point[i] = read();
for(int i = 1; i <= m; i++) {
x[i] = read(), y[i] = read();
add(x[i], y[i]);
}
for(int i = 1; i <= n; i++)
if(!dfn[i])
tarjan(i);
for(int i = 1; i <= m; i++)//缩点后重新建边。
if(scc[x[i]] != scc[y[i]]) {
in[scc[y[i]]]++;
to1[cnt1] = scc[y[i]];
nex1[cnt1] = head1[scc[x[i]]];
head1[scc[x[i]]] = cnt1++;
}
bfs();
return 0;
}
Popular Cows
思路
显然是一道缩点的题目，缩点完后，我们可以知道如果一个强连通分量的出度为零，并且只有一个强连通分量的初读为零，那么缩点后的图一定时联通的，这个时候出度为零的强连通分量重的点的个数就是我们要求的答案。
代码
// #include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
inline ll read() {
ll f = 1, x = 0;
char c = getchar();
while(c > '9' || c < '0') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f * x;
}
const int N1 = 1e4 + 10, N2 = 5e4 + 10;
int head[N1], to[N2], nex[N2], cnt = 1;
int visit[N1], dfn[N1], low[N1], scc[N1], sz[N1], n, m, tot, sum;
int x[N2], y[N2], out[N1];
int stk[N1], top;
void add(int x, int y) {
to[cnt] = y;
nex[cnt] = head[x];
head[x] = cnt++;
}
void tarjan(int rt) {
visit[rt] = 1, stk[++top] = rt;
dfn[rt] = low[rt] = ++tot;
for(int i = head[rt]; i; i = nex[i]) {
if(!dfn[to[i]]) {
tarjan(to[i]);
low[rt] = min(low[rt], low[to[i]]);
}
else if(visit[to[i]])
low[rt] = min(low[rt], dfn[to[i]]);
}
if(dfn[rt] == low[rt]) {
sum++;
do {
scc[stk[top]] = sum;
sz[sum]++;
visit[stk[top]] = 0;
top--;
}while(rt != stk[top + 1]);
}
}
int main() {
// freopen("in.txt", "r", stdin);
while(scanf("%d %d", &n, &m) != EOF) {
for(int i = 1; i <= n; i++)
head[i] = visit[i] = sz[i] = out[i] = dfn[i] = 0;
cnt = 1, tot = sum = top = 0;
for(int i = 1; i <= m; i++) {
x[i] = read(), y[i] = read();
add(x[i], y[i]);
}
for(int i = 1; i <= n; i++)
if(!dfn[i])
tarjan(i);
// puts("okkkkk");
for(int i = 1; i <= m; i++)
if(scc[x[i]] != scc[y[i]])
out[scc[x[i]]]++;
int num = 0, ans = 0;
for(int i = 1; i <= sum; i++)
if(out[i] == 0) {
ans = sz[i];
num++;
}
if(num != 1) puts("0");
else printf("%d\n", ans);
}
return 0;
}
Bomb
思路
容易想到爆炸就是一个传递的图，当爆炸形成一个环的时候，明显可以进行缩点操作，所以当我们进行完缩点之后，我们只要统计剩余的点中入度为零的点就行，同时我们需要的花费就是这些点所在的联通分量中的花费最小的点。
代码
#include
using namespace std;
typedef long long ll;
ll read() {
ll f = 1, x = 0;
char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f * x;
}
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N1 = 1e3 + 10, N2 = 1e6 + 10;
int head[N1], to[N2], nex[N2], cnt;
int visit[N1], dfn[N1], low[N1], scc[N1], n, sum, tot;
int stk[N1], in[N1], top;
ll cost[N1];
struct point {
ll x, y, c, r;
void input() {
x = read(), y = read(), r = read(), c = read();
}
}a[N1];
ll dis(point a, point b) {
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
void add(int x, int y) {
to[cnt] = y;
nex[cnt] = head[x];
head[x] = cnt++;
}
void tarjan(int rt) {
dfn[rt] = low[rt] = ++tot;
stk[++top] = rt;
visit[rt] = 1;
for(int i = head[rt]; i; i = nex[i]) {
if(!dfn[to[i]]) {
tarjan(to[i]);
low[rt] = min(low[rt], low[to[i]]);
}
else if(visit[to[i]])
low[rt] = min(low[rt], dfn[to[i]]);
}
if(dfn[rt] == low[rt]) {
sum++;
do {
scc[stk[top]] = sum;
cost[sum] = min(cost[sum], a[stk[top]].c);
visit[top[stk]] = 0;
top--;
}while(stk[top + 1] != rt);
}
}
int main() {
// freopen("in.txt", "r", stdin);
int t = read();
for(int cas = 1; cas <= t; cas++) {
n = read();
for(int i = 1; i <= n; i++) {
head[i] = visit[i] = dfn[i] = low[i] = scc[i] = in[i] = 0;
cost[i] = INF;
a[i].input();
}
tot = sum = top = 0, cnt = 1;
for(int i = 1; i <= n; i++)
for(int j = i + 1; j <= n; j++) {
ll d = dis(a[i], a[j]);
if(d <= a[i].r * a[i].r)
add(i, j);
if(d <= a[j].r * a[j].r)
add(j, i);
}
for(int i = 1; i <= n; i++)
if(!dfn[i])
tarjan(i);
for(int i = 1; i <= n; i++)
for(int j = head[i]; j; j = nex[j])
if(scc[i] != scc[to[j]])
in[scc[to[j]]]++;
ll ans = 0;
for(int i = 1; i <= sum; i++)
if(in[i] == 0)
ans += cost[i];
printf("Case #%d: %lld\n", cas, ans);
}
return 0;
}