scu图论专题题解

传送门：点击打开链接

感觉做着还是好费劲啊，，，有些东西总是想不到T^T

A题：给出边，分别判断是有向图或者无向图的时候，是否为欧拉回路

首先必须弱连通图要连通。对于无向图，度全为偶数，或者度为奇数的点的个数为2

对于有向图，入度全部等于出度，或者1个点入度-出度=1，1个点出度-入度=1，其他点入度等于出度

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 1000 + 5;int P[MX], IN[MX], OUT[MX];int find(int x) {    return P[x] == x ? x : (P[x] = find(P[x]));}int main() {    int T, n, m; //FIN;    scanf("%d", &T);    while(T--) {        scanf("%d%d", &n, &m);        memset(IN, 0, sizeof(IN));        memset(OUT, 0, sizeof(OUT));        for(int i = 1; i <= n; i++) P[i] = i;        for(int i = 1; i <= m; i++) {            int u, v;            scanf("%d%d", &u, &v);            IN[v]++; OUT[u]++;            u = find(u); v = find(v);            P[v] = u;        }        bool a1 = 0, a2 = 0;        int c1 = 0, c2 = 0, c3 = 0;        for(int i = 1; i <= n; i++) {            if((IN[i] + OUT[i]) % 2) c1++;        }        if(c1 == 2 || c1 == 0) a1 = 1;        c1 = 0;        for(int i = 1; i <= n; i++) {            if(IN[i] - OUT[i] == 1) c1++;            if(IN[i] - OUT[i] == -1) c2++;            if(IN[i] == OUT[i]) c3++;        }        if(c3 == n || (c3 == n - 2 && c1 == 1 && c2 == 1)) a2 = 1;        c1 = 0;        for(int i = 1; i <= n; i++) {            if(i == find(i)) c1++;        }        if(c1 != 1) a1 = a2 = 0;        printf("%s %s\n", a1 ? "Yes" : "No", a2 ? "Yes" : "No");    }    return 0;}

B题：

我猜标程是想考，如果覆盖，就建一条边，那么最后只要看是否能做拓扑排序，就有答案了。

不过这题也有暴力方法，因为数据比较小，每次我都删掉最上面的那个窗口，直到全部删掉即可

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 10 + 5;int s[MX][MX];inline bool ok(int i, int j, int w) {    if(s[i][j] == 0 || s[i][j] == w) return 1;    return 0;}bool solve() {    while(1) {        int tot = 0, flag = 0;        for(int i = 1; i <= 3; i++) {            for(int j = 1; j <= 3; j++) {                tot++;                if(ok(i, j, tot) && ok(i + 1, j, tot) && ok(i, j + 1, tot) && ok(i + 1, j + 1, tot)) {                    int sum = s[i][j] + s[i + 1][j] + s[i][j + 1] + s[i + 1][j + 1];                    if(sum) s[i][j] = s[i + 1][j] = s[i][j + 1] = s[i + 1][j + 1] = 0, flag = 1;                }            }        }        if(!flag) break;    }    int w = 0;    for(int i = 1; i <= 4; i++) {        for(int j = 1; j <= 4; j++) {            w += s[i][j];        }    }    return w == 0;}int main() {    int T;//FIN;    scanf("%d", &T);    while(T--) {        for(int i = 1; i <= 4; i++) {            for(int j = 1; j <= 4; j++) {                scanf("%d", &s[i][j]);            }        }        printf("%s\n", solve() ? "Lucky dog!" : "BOOM!");    }    return 0;}

C题：对于任意两个字符，如果ASCII码之差绝对值<=1,则就会连一条边。只有a，b，c三种字符

现在告诉你一个图，问是否有这样的满足题意的字符集合。

当一个点连着所有其他的点的话，我们就让这个点的字符为b，这样一定是最优的

那么对于a和c，很明显是一个二分图，我们接下来就类似判定图是否是二分图的方法，01染色即可

不过要注意，如果i到j有一条边，j点不是字符b，那么j的字符必须和i的字符一样

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 500 + 5;int n, m, w[MX];bool G[MX][MX], vis[MX];bool DFS(int u, int x) {    vis[u] = 1; w[u] = x;    for(int v = 1; v <= n; v++) {        if(u == v) continue;        if(G[u][v]) {            if(vis[v]) {                if(w[v] == (x ^ 1)) return 0;            } else if(!DFS(v, x)) return 0;        } else {            if(vis[v]) {                if(w[v] != (x ^ 1)) return 0;            } else if(!DFS(v, x ^ 1)) return 0;        }    }    return 1;}bool solve() {    for(int i = 1; i <= n; i++) {        if(!vis[i] && !DFS(i, 0)) return 0;    }    return 1;}int main() {    //FIN;    int T; scanf("%d", &T);    while(T--) {        scanf("%d%d", &n, &m);        memset(G, 0, sizeof(G));        memset(vis, 0, sizeof(vis));        for(int i = 1; i <= m; i++) {            int u, v;            scanf("%d%d", &u, &v);            G[u][v] = G[v][u] = 1;        }        for(int i = 1; i <= n; i++) {            int cnt = 0;            for(int j = 1; j <= n; j++) {                cnt += G[i][j];            }            if(cnt == n - 1) {                vis[i] = 1; w[i] = 2;            }        }        printf("%s\n", solve() ? "Yes" : "No");    }    return 0;}

D:告诉所有人的坐标，和2个食堂的坐标。有的人互相喜欢，必须要去同一个食堂，有的人互相讨厌，不能在一个食堂。要求任意两个人所走的距离+两个人选择的食堂直接的距离之和最小。

一个人有2种状态，要么1食堂，要么2食堂，我们会比较容易想到2sat。那么假如我们直接二分答案，对于前面的讨厌和喜欢的，我们很容易就能连出边。

之后我们2个for循环，枚举任意一对人，然后再枚举他们分别要去哪个食堂，看这种情况是否符合要求，如果不符合，就连边让这种情况禁止

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 6000 + 5;const int INF = 0x3f3f3f3f;struct Edge {    int v, nxt;} E[5000005];int Head[MX][2], erear;void edge_init() {    erear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int z, int u, int v) {    E[erear].v = v;    E[erear].nxt = Head[u][z];    Head[u][z] = erear++;}void edge_add(int u, int v) {    edge_add(0, u, v);    edge_add(1, v, u);}int Stack[MX], Belong[MX], vis[MX], ssz, bsz;void DFS(int u, int s) {    vis[u] = 1;    if(s) Belong[u] = s;    for(int i = Head[u][s > 0]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(!vis[v]) DFS(v, s);    }    if(!s) Stack[++ssz] = u;}void tarjan(int n) {    ssz = bsz = 0;    for(int i = 1; i <= n; i++) vis[i] = 0;    for(int i = 1; i <= n; i++) {        if(!vis[i]) DFS(i, 0);    }    for(int i = 1; i <= n; i++) vis[i] = 0;    for(int i = ssz; i >= 1; i--) {        if(!vis[Stack[i]]) DFS(Stack[i], ++bsz);    }}int n, A, B;struct Point {    int x, y;} P[MX], din[2];int lku[MX], lkv[MX];int ulku[MX], ulkv[MX];int dist(Point a, Point b) {    return abs(a.x - b.x) + abs(a.y - b.y);}bool check(int m) {    edge_init();    for(int i = 1; i <= A; i++) {        edge_add(ulku[i], ulkv[i] + n); edge_add(ulku[i] + n, ulkv[i]);        edge_add(ulkv[i], ulku[i] + n); edge_add(ulkv[i] + n, ulku[i]);    }    for(int i = 1; i <= B; i++) {        edge_add(lku[i], lkv[i]); edge_add(lkv[i], lku[i]);        edge_add(lku[i] + n, lkv[i] + n); edge_add(lkv[i] + n, lku[i] + n);    }    for(int i = 1; i <= n; i++) {        for(int j = i + 1; j <= n; j++) {            for(int a = 0; a <= 1; a++) {                for(int b = 0; b <= 1; b++) {                    if(dist(P[i], din[a]) + dist(P[j], din[b]) + dist(din[a], din[b]) > m) {                        edge_add(!a ? i : i + n, !b ? j + n : j);                        edge_add(!b ? j : j + n, !a ? i + n : i);                    }                }            }        }    }    tarjan(2 * n);    for(int i = 1; i <= n; i++) {        if(Belong[i] == Belong[i + n]) return false;    }    return true;}int solve() {    int l = 0, r = 1e7, m;    if(!check(r)) return -1;    while(l <= r) {        m = (l + r) >> 1;        if(check(m)) r = m - 1;        else l = m + 1;    }    return r + 1;}int main() {    int T;//FIN;    scanf("%d", &T);    while(T--) {        scanf("%d%d%d", &n, &A, &B);        scanf("%d%d%d%d", &din[0].x, &din[0].y, &din[1].x, &din[1].y);        for(int i = 1; i <= n; i++) {            scanf("%d%d", &P[i].x, &P[i].y);        }        for(int i = 1; i <= A; i++) {            scanf("%d%d", &ulku[i], &ulkv[i]);        }        for(int i = 1; i <= B; i++) {            scanf("%d%d", &lku[i], &lkv[i]);        }        printf("%d\n", solve());    }    return 0;}

E题：敢写就敢过啊。。

先跑强连通分量，之后我们能很明显的发现，如果在一个组的话，那么这个组里面的拓扑序就是唯一的。我们很容易就想到了二分图的最小路径覆盖。

不过复杂度如果在极限情况下，应该是蛮差的，如果不敢用匈牙利算法，那就写更快的，虽然我不会写

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int> PII;const int MX = 1e4 + 5;const int INF = 0x3f3f3f3f;struct Edge {    int u, v, nxt;} E[200005];int Head[MX], erear;void edge_init() {    erear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v) {    E[erear].u = u;    E[erear].v = v;    E[erear].nxt = Head[u];    Head[u] = erear++;}int n, m, IN[MX], cnt[MX], val[MX];int bsz, ssz, dsz;int Low[MX], DFN[MX];int belong[MX], Stack[MX];bool inStack[MX];void Init_tarjan(int n) {    bsz = ssz = dsz = 0;    for(int i = 1; i <= n; ++i) Low[i] = DFN[i] = 0;}void Tarjan(int u) {    Stack[++ssz] = u;    inStack[u] = 1;    Low[u] = DFN[u] = ++dsz;    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(!DFN[v]) {            Tarjan(v);            Low[u] = min( Low[v], Low[u]);        } else if(inStack[v]) {            Low[u] = min( Low[u], DFN[v]);        }    }    if(Low[u] == DFN[u]) {        ++bsz;        int v;        do {            v = Stack[ssz--];            inStack[v] = 0;            belong[v] = bsz;        } while(u != v);    }}int match[MX];bool vis[MX];bool DFS(int u) {    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(!vis[v]) {            vis[v] = 1;            if(match[v] == -1 || DFS(match[v])) {                match[v] = u;                return 1;            }        }    }    return 0;}int BM(int n) {    int res = 0;    memset(match, -1, sizeof(match));    for(int u = 1; u <= n; u++) {        memset(vis, 0, sizeof(vis));        if(DFS(u)) res++;    }    return res;}int solve(int n) {    Init_tarjan(n);    for (int i = 1; i <= n; i++) {        if (!DFN[i]) Tarjan(i);    }    edge_init();    for(int i = 0; i < m; i++) {        int u = E[i].u, v = E[i].v;        u = belong[u]; v = belong[v];        if(u != v) edge_add(u, v + bsz);    }    return bsz - BM(bsz);}int main() {    int T; //FIN;    scanf("%d", &T);    while(T--) {        edge_init();        scanf("%d%d", &n, &m);        for(int i = 1; i <= m; i++) {            int u, v;            scanf("%d%d", &u, &v);            edge_add(u, v);        }        printf("%d\n", solve(n));    }    return 0;}

F题:令x为一个环中的边的最大值，那么有多少个环，我们就得到了多少个x。答案要求这些x中的最小的那个

这题非常有意思啊，通常看到环，我们可能会想到强连通分量，不过这个要求所有的环，强连通是做不到的。

我们先求出一个最小生成树。然后我们再枚举所有的边。

如果这条边是生成树上的，我们就忽略这条边，因为这条边不能组成环

如果这条边不是生成树上的，那么我们把u和v连起来，就会和树上的路径形成环。又因为是最小生成树，这条边的权值一定会大于路径上任意一条边的权值，所以对于这个环，边长的最大值就等于我们枚举到的这条边的权值。

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 5e5 + 5;const int INF = 0x3f3f3f3f;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1struct Edge {    int u, v, cost;    bool operator<(const Edge &P) const {        return cost < P.cost;    }} A[2000005];bool ok[MX];int n, m, P[MX];int find(int x) {    return P[x] == x ? x : (P[x] = find(P[x]));}void MST_solve() {    sort(A + 1, A + 1 + m);    for(int i = 1; i <= n; i++) P[i] = i;    for(int i = 1; i <= m; i++) {        int p1 = find(A[i].u), p2 = find(A[i].v);        if(p1 != p2) {            P[p1] = p2;            ok[i] = 1;        }    }}int main() {    int T;//FIN;    scanf("%d", &T);    while(T--) {        scanf("%d%d", &n, &m);        memset(ok, 0, sizeof(ok));        for(int i = 1; i <= m; i++) {            scanf("%d%d%d", &A[i].u, &A[i].v, &A[i].cost);        }        MST_solve();        int ans = INF;        for(int i = 1; i <= m; i++) {            if(!ok[i]) ans = min(ans, A[i].cost);        }        if(ans == INF) printf("No solution!\n");        else printf("%d\n", ans);    }    return 0;}

G题：我们先求出哪些边会构成最短路。只要从起点出发求一次，从终点出发求一次，然后枚举边，看从起点到u的最短距离+边长+终点到v的最短距离=起点到终点的最短距离，那么这条边就是能构成最短路的。我们得到后来这个DAG模型后，假设起点有INF条流，然后分配给下面的节点，类似网络流的思想，只是没有容量的概念，所以胡乱向下更新就行了。

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 1e3 + 5;const int MS = 2e5 + 5;const LL INF = 0x3f3f3f3f3f3f3f3fLL;struct Edge {    bool ok, vis;    int u, v, w, nxt;} E[MS];int Head[MX], erear;void edge_init() {    erear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v, int w) {    E[erear].u = u;    E[erear].v = v;    E[erear].w = w;    E[erear].ok = E[erear].vis = 0;    E[erear].nxt = Head[u];    Head[u] = erear++;}int n, m, op, ed;int val[MX];LL d[2][MX], path;typedef pair<LL, int> PLI;void dijkstra(int u, LL d[], int o) {    for(int i = 1; i <= n; i++) d[i] = INF;    priority_queue<PLI, vector<PLI>, greater<PLI> > Q;    d[u] = 0; Q.push(PLI(d[u], u));    while(!Q.empty()) {        PLI ftp = Q.top(); Q.pop();        LL dist = ftp.first; int u = ftp.second;        if(dist > d[u]) continue;        for(int i = Head[u]; ~i; i = E[i].nxt) {            if((i & 1) != o) continue;            int v = E[i].v, w = E[i].w;            if(d[u] + w < d[v]) {                d[v] = d[u] + w;                Q.push(PLI(d[v], v));            }        }    }}int solve() {    queue<int> Q;    memset(val, 0, sizeof(val));    Q.push(op); val[op] = INF;    while(!Q.empty()) {        int u = Q.front(); Q.pop();        for(int i = Head[u]; ~i; i = E[i].nxt) {            if(!E[i].ok || E[i].vis || !val[u]) continue;            E[i].vis = 1; val[u]--;            Q.push(E[i].v); val[E[i].v]++;        }    }    return val[ed];}int main() {    //FIN;    int T; scanf("%d", &T);    while(T--) {        edge_init();        scanf("%d%d", &n, &m);        for(int i = 1; i <= m; i++) {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            edge_add(u, v, w);            edge_add(v, u, w);        }        scanf("%d%d", &op, &ed);        dijkstra(op, d[0], 0);        path = d[0][ed];        dijkstra(ed, d[1], 1);        for(int i = 0; i < erear; i += 2) {            int u = E[i].u, v = E[i].v, w = E[i].w;            if(d[0][u] + w + d[1][v] == path) {                E[i].ok = 1;            }        }        printf("%d\n", solve());    }    return 0;}

H题：被题目迷惑了。。

冷静的想一下，就是要从起点到终点，所花费的时间<=k,并且经过的路径里的cap的最小值最大。

这种让什么什么最大的题，先无脑二分想一想，发现好像非常有道理的样子。。我们直接二分答案，之后跑logn次最短路就可以了

不过我第一次TLE了，再在最短路里加个特判，如果此时路径距离已经>k了，就直接返回

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int> PII;typedef pair<LL, int>PLI;const int MX = 1e5 + 5;const LL INF = 0x3f3f3f3f3f3f3f3fLL;struct Edge {    int v, nxt, w, cap;} E[MX];int Head[MX], erear;void edge_init() {    erear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v, int cap, int w) {    E[erear].v = v;    E[erear].w = w;    E[erear].cap = cap;    E[erear].nxt = Head[u];    Head[u] = erear++;}int n, m, k;LL d[MX];LL dijkstra(int u, int m) {    for(int i = 1; i <= n; i++) d[i] = INF;    priority_queue<PLI, vector<PLI>, greater<PLI> >Q;    d[u] = 0; Q.push(PLI(0, u));    while(!Q.empty()) {        PLI ftp = Q.top(); Q.pop();        int u = ftp.second;        LL dist = ftp.first;        if(dist > d[u]) continue;        if(dist > k) return d[n];        for(int i = Head[u]; ~i; i = E[i].nxt) {            int v = E[i].v, w = E[i].w, cap = E[i].cap;            if(cap >= m && d[u] + w < d[v]) {                d[v] = d[u] + w;                Q.push(PLI(d[v], v));            }        }    }    return d[n];}bool check(int m) {    return dijkstra(1, m) <= k;}int main() {    int T;//FIN;    scanf("%d", &T);    while(T--) {        edge_init();        scanf("%d%d%d", &n, &m, &k);        LL l = 2e9, r = 1, mid;        for(int i = 1; i <= m; i++) {            int u, v, cap, dis;            scanf("%d%d%d%d", &u, &v, &cap, &dis);            edge_add(u, v, cap, dis);            edge_add(v, u, cap, dis);            l = min(l, (LL)cap);            r = max(r, (LL)cap);        }        if(!check(l)) {            printf("Poor RunningPhoton!\n");            continue;        }        while(l <= r) {            mid = (l + r) >> 1;            if(check(mid)) l = mid + 1;            else r = mid - 1;        }        printf("%lld\n", l - 1);    }    return 0;}

I题：tarjan求割边

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 2e5 + 5;const int INF = 0x3f3f3f3f;int rear;int Head[MX], Next[MX];int Low[MX], DFN[MX], dfs_clock;int cut[MX];struct Edge {    int u, v, sign;} E[MX];void edge_init() {    rear = 0;    memset(Head, -1, sizeof(Head));    memset(Next, -1, sizeof(Next));}void edge_add(int u, int v) {    E[rear].u = u;    E[rear].v = v;    E[rear].sign = false;    Next[rear] = Head[u];    Head[u] = rear++;}void tarjan_init() {    dfs_clock = 0;    memset(DFN, 0, sizeof(DFN));    memset(cut, 0, sizeof(cut));}int tarjan(int u, int e) {    Low[u] = DFN[u] = ++dfs_clock;    int child = 0;    for(int id = Head[u]; ~id; id = Next[id]) {        int v = E[id].v;        if(!DFN[v]) {            int lowv = tarjan(v, id | 1);            Low[u] = min(Low[u], lowv);            if(lowv >= DFN[u]) {                cut[u] = 1;            }            if(lowv > DFN[u]) {                E[id].sign = 1;                E[id ^ 1].sign = 1;            }            child++;        } else if((id | 1) != e && DFN[v] < DFN[u]) {            Low[u] = min(Low[u], DFN[v]);        }    }    if(e == -1 && child == 1) cut[u] = 0;    return Low[u];}int main() {    //FIN;    int T, ansk = 0;    scanf("%d", &T);    while(T--) {        int n, m;        scanf("%d%d", &n, &m);        edge_init();        tarjan_init();        for(int i = 1; i <= m; i++) {            int u, v;            scanf("%d%d", &u, &v);            edge_add(u, v);            edge_add(v, u);        }        for(int i = 1; i <= n; i++) {            if(!DFN[i]) tarjan(i, -1);        }        int ans = 0;        for(int i = 0; i < rear; i += 2) {            if(E[i].sign) ans++;        }        printf("Case %d: %d\n", ++ansk, ans);    }    return 0;}

J题：首先我们求一遍强连通分量然后缩点，然后就变成了DAG模型。

然后我们能发现，这也是二分图的最小路径覆盖中的经典题。不过这里边是可以重复经过的，我们需要多增加许多条边。如果u能通过有向边到v，那么我们就在u连一条边到v。这一部分我们直接用bitset然后记忆化搜索，然后再建边，复杂度只有O(n^2)，比floyd稍微快一些，而且还蛮好用的。

理论上这题的复杂度用BM也是通过不了的，不过敢写敢过啊，过不了再换更快的求二分图匹配的写法啊，嘿嘿嘿

#include <map>#include <set>#include <cmath>#include <ctime>#include <stack>#include <queue>#include <cstdio>#include <cctype>#include <bitset>#include <string>#include <vector>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define fuck(x) cout<<"["<<x<<"]";#define FIN freopen("input.txt","r",stdin);#define FOUT freopen("output.txt","w+",stdout);//#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;typedef long long LL;typedef pair<int, int>PII;const int MX = 2e3 + 5;const int INF = 0x3f3f3f3f;struct Edge {    int u, v, nxt;} E[200005];int Head[MX], erear;void edge_init() {    erear = 0;    memset(Head, -1, sizeof(Head));}void edge_add(int u, int v) {    E[erear].u = u;    E[erear].v = v;    E[erear].nxt = Head[u];    Head[u] = erear++;}int bsz, ssz, dsz;int Low[MX], DFN[MX];int belong[MX], Stack[MX];bool inStack[MX];void Init_tarjan(int n) {    bsz = ssz = dsz = 0;    for(int i = 1; i <= n; ++i) Low[i] = DFN[i] = 0;}void Tarjan(int u) {    Stack[++ssz] = u;    inStack[u] = 1;    Low[u] = DFN[u] = ++dsz;    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(!DFN[v]) {            Tarjan(v);            Low[u] = min(Low[v], Low[u]);        } else if(inStack[v]) {            Low[u] = min(Low[u], DFN[v]);        }    }    if(Low[u] == DFN[u]) {        ++bsz;        int v;        do {            v = Stack[ssz--];            inStack[v] = 0;            belong[v] = bsz;        } while(u != v);    }}int n, m;int match[MX];bool vis[MX];typedef bitset<1000> bits;bits G[MX];bool DFS(int u) {    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        if(!vis[v]) {            vis[v] = 1;            if(match[v] == -1 || DFS(match[v])) {                match[v] = u;                return 1;            }        }    }    return 0;}int BM(int n) {    int res = 0;    memset(match, -1, sizeof(match));    for(int u = 1; u <= n; u++) {        memset(vis, 0, sizeof(vis));        if(DFS(u)) res++;    }    return res;}bits DP(int u) {    if(G[u].count()) return G[u];    G[u][u - 1] = 1;    for(int i = Head[u]; ~i; i = E[i].nxt) {        int v = E[i].v;        G[u] |= DP(v);    }    return G[u];}int solve(int n) {    Init_tarjan(n);    for (int i = 1; i <= n; i++) {        if (!DFN[i]) Tarjan(i);    }    edge_init();    for(int i = 0; i < m; i++) {        int u = E[i].u, v = E[i].v;        u = belong[u]; v = belong[v];        if(u != v) edge_add(u, v);    }    for(int i = 1; i <= bsz; i++) G[i].reset();    for(int i = 1; i <= bsz; i++) DP(i);    edge_init();    for(int i = 1; i <= bsz; i++) {        for(int j = 1; j <= bsz; j++) {            if(i != j && G[i][j - 1]) edge_add(i, j + bsz);        }    }    return bsz - BM(bsz);}int main() {    int T, ansk = 0; //FIN;    scanf("%d", &T);    while(T--) {        edge_init();        scanf("%d%d", &n, &m);        for(int i = 1; i <= m; i++) {            int u, v;            scanf("%d%d", &u, &v);            edge_add(u, v);        }        printf("Case %d: %d\n", ++ansk, solve(n));    }    return 0;}

K题：应该就是求强连通分量缩点，然后在DAG模型上按拓扑序乱搞一下。

不是特别懂题目意思，，不写了- -

L题：bnu校赛原题，题解点击打开链接

M题：不会，感觉是数论，但是Claris直接秒了，而且是图论，不是很懂，以后再补。。