Description
AGC027C
给定一张图,点有标号A或B,计算是否对于任意的一个由AB构成的字符串都在图中有对应的路径。
Solution
观察可以发现,如果有个环是AABB的无限循环,就输出Yes
,否则输出No
。
然鹅,话是这么说,码却没有那么好写,我一开始写了dfs
找环,但是一直WA,和标程对拍也拍不出来错。所以我就换了一种思路,借用拓扑排序的思想,每次将之与一种字符相连的点删去,如果删完还剩下,那么就说明有一个AABB的环,否则就没有。
Code
拓扑排序版:
#include <cstdio>
const int N = 2e5 + 10;
const int M = 4e5 + 10;
int hd[N], nxt[M], to[M], cnt;
int deg[N][2];
int n, m;
char col[N];
int q[N], qhd, qtl, vis[N];
inline void adde(int x, int y) {
cnt++;
to[cnt] = y; nxt[cnt] = hd[x];
hd[x] = cnt;
}
int main() {
scanf("%d%d", &n, &m);
scanf("%s", col+1);
for (int i = 1, x, y; i <= m; ++i) {
scanf("%d%d", &x, &y);
adde(x, y);
adde(y, x);
deg[x][col[y] == 'B']++;
deg[y][col[x] == 'B']++;
}
for (int i = 1; i <= n; ++i) {
if (deg[i][0] == 0 || deg[i][1] == 0) vis[q[qtl++] = i] = 1;
}
while (qhd != qtl) {
int x = q[qhd++];
// printf("%d\n", x);
// vis[x] = 1;
for (int i = hd[x]; i; i = nxt[i]) if (!vis[to[i]]) {
if (--deg[to[i]][col[x] == 'B'] == 0) {
vis[to[i]] = 1;
q[qtl++] = to[i];
}
}
}
if (qtl == n) puts("No");
else puts("Yes");
}
并不知道有什么错WA个不停的dfs版:
#include <cstdio>
const int N = 2e5 + 10;
const int M = 4e5 + 10;
const char S[] = "ABBAA";
int hd[N], nxt[M], to[M], cnt;
char col[N];
int n, m, flag, pos;
inline void adde(int x, int y) {
cnt++;
nxt[cnt] = hd[x];
to[cnt] = y;
hd[x] = cnt;
}
inline void dfs(int x, int c, int nf) {
if (flag) return;
if (c == 0 && nf) {
if (x == pos) flag = 1;
return;
}
for (int i = hd[x]; i; i = nxt[i]) {
if (col[to[i]] == S[c])
dfs(to[i], (c+1)%4, 1);
}
}
int main() {
scanf("%d%d", &n, &m);
scanf("%s", col+1);
for (int i = 1, x, y; i <= m; ++i) {
scanf("%d%d", &x, &y);
adde(x, y);
adde(y, x);
}
for (int i = 1; i <= n; ++i) {
if (col[i] == 'A') {
pos = i;
dfs(i, 0, 0);
}
if (flag) break;
}
printf(flag ? "Yes\n" : "No\n");
return 0;
}
求查错或者给出一组数据卡掉这个程序。