标签：c++ max ack ems 判断 queue 关系 min 选择
题意和思路看这篇博客就行了：https://www.cnblogs.com/cjyyb/p/10507937.html
有个问题需要注意：对于每个scc，只需要考虑进入这个scc的时间即可，其实和从哪个点进没有关系，因为scc内每个点都可以互相到达，所以只需记录时间就囊括了所有的情况，比如时间3从1号点进和时间4从2号点进是等价的，这也是为什么可以随便选择一颗生成树的原因。对于scc的出边，边的长度是val[u] + val[v] - 1,因为假设从scc x的根 到点scc y的点v，时间是val[u] + 1,而scc y的根到v的距离是val[v]，所以要求原装态需要减去这个。当然，这个边需要对两个scc的gcd的gcd取模，以判断从这条边到下一条边可以到达的状态。
代码：
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int head[maxn], Next[maxn * 2], ver[maxn * 2], tot;
int heade[maxn], Nexte[maxn * 2 * 50], edgee[maxn * 2 * 50], vere[maxn * 2 * 50], tote, cnt;
int n, m, D, GCD, g[maxn], f[maxn][51], pre[maxn][51], deg[maxn], val[maxn], c[maxn];
int dfn[maxn], low[maxn], num, top, Stack[maxn];
char v[maxn][51];
bool vis[maxn], ins[maxn];
vector<int> scc[maxn];
void add(int x, int y) {
ver[++tot] = y;Next[tot] = head[x];head[x] = tot;
}
void adde(int x, int y, int z) {
vere[++tote] = y, edgee[tote] = z, Nexte[tote] = heade[x], heade[x] = tote, deg[y]++;
}
void solve(int x, int p) {
val[x] = p, vis[x] = 1;
for (int i = head[x]; i; i = Next[i]) {
int y = ver[i];
if(c[y] != cnt) continue;
if(!vis[y]) solve(y, p + 1);
else GCD = __gcd(GCD, abs(p + 1 - val[y]));
}
}
void tarjan(int x) {
dfn[x] = low[x] = ++num;
Stack[++top] = x, ins[x] = 1;
for (int i = head[x]; i; i = Next[i]) {
if (!dfn[ver[i]]) {
tarjan(ver[i]);
low[x] = min(low[x], low[ver[i]]);
} else if (ins[ver[i]])
low[x] = min(low[x], dfn[ver[i]]);
}
if (dfn[x] == low[x]) {
cnt++;
int y;
do {
y = Stack[top--], ins[y] = 0;
c[y] = cnt, scc[cnt].push_back(y);
} while(x != y);
GCD = D, solve(x, 0); g[cnt] = GCD;
for (int i = 0; i < GCD; i++)
for (int j = 0; j < scc[cnt].size(); j++) {
for (int k = i; k < D; k += GCD)
if(v[scc[cnt][j]][k] == ‘1‘) {
v[scc[cnt][j]][i] = ‘1‘;
break;
}
}
for (int i = 0; i < GCD; i++)
for (int j = 0; j < scc[cnt].size(); j++) {
if(v[scc[cnt][j]][(val[scc[cnt][j]] + i) % GCD] == ‘1‘)
pre[cnt][i]++;
}
}
}
void dp(void) {
int ans = 0;
memset(f, -0x3f, sizeof(f));
f[c[1]][0] = pre[c[1]][0];
queue<int> q;
for (int i = 1; i <= cnt; i++) {
if (!deg[i])
q.push(i);
}
while(!q.empty()) {
int x = q.front();
q.pop();
for (int i = heade[x]; i; i = Nexte[i]) {
int y = vere[i];
deg[y]--;
if(deg[y] == 0)
q.push(y);
}
for (int i = heade[x]; i; i = Nexte[i]) {
for (int j = 0; j < D; j++) {
int y = vere[i], z = edgee[i];
f[y][(z + j) % D] = max(f[y][(z + j) % D], f[x][j] + pre[y][(z + j) % g[y]]);
}
}
for (int i = 0; i < D; i++)
ans = max(ans, f[x][i]);
}
printf("%d\n", ans);
}
int main() {
int x, y;
scanf("%d%d%d", &n, &m, &D);
for (int i = 1; i <= m; i++) {
scanf("%d%d", &x, &y);
add(x, y);
}
for (int i = 1; i <= n; i++)
scanf("%s", v[i]);
for (int i = 1; i <= n; i++) {
if(!dfn[i])
tarjan(i);
}
for (int j = 1; j <= n; j++)
for (int i = head[j]; i; i = Next[i]) {
int y = ver[i];
if(c[j] == c[y]) continue;
int dd = __gcd(g[c[j]], g[c[y]]);
int d = (val[j] - val[y] + 1 + D) % D;
for (int k = d % dd; k < D; k += dd) {
adde(c[j], c[y], k);
}
}
dp();
}
Codeforces 1137C Museums Tour （强连通分量， DP）
标签：c++ max ack ems 判断 queue 关系 min 选择
原文地址：https://www.cnblogs.com/pkgunboat/p/10562413.html
