ZOJ 3795 - Grouping (强连通+dp)
2015-07-20 16:13
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题目:
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5303
题意:
N个数,M对数之间的关系,。没有关系的数才能放在同一个集合,求出最少的集合数。
思路:
题目转化为求有向图中节点数最大的节点数集,使得节点数集中任意两个点u和v满足,u可以到达v或者是v可以到达u。(因为互相有关系的点必须放在不同的集合中)
将数之间的关系建有向图,跑 强连通。缩点 得到强连通图。
在强连通图上 dp,求出最长的连续序列。
做过类似的题目:http://blog.csdn.net/sotifish/article/details/44681225
AC.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
using namespace std;
const int MAXN = 1e5+100;
const int MAXM = 3e5+100;
int N, M;
struct Edge {
int to, next;
}edge[MAXM];
int head[MAXN], tot;
int low[MAXN], dfn[MAXN], stak[MAXN], bel[MAXN], num[MAXN];
int idx, top, scc;
bool instk[MAXN];
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void tarjan(int u)
{
int v;
low[u] = dfn[u] = ++idx;
stak[top++] = u;
instk[u] = true;
for(int i = head[u]; ~i; i = edge[i].next) {
v = edge[i].to;
if(!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(instk[v]) {
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u]) {
scc++;
do {
v = stak[--top];
instk[v] = false;
bel[v] = scc;
num[scc]++;
}while(v!=u);
}
}
void Scc()
{
memset(dfn, 0, sizeof(dfn));
memset(instk, 0, sizeof(instk));
memset(num, 0, sizeof(num));
idx = scc = top = 0;
for(int i = 1; i <= N; ++i) {
if(!dfn[i]) {
tarjan(i);
}
}
}
vector<int> G[MAXN];
int dp[MAXN];
void deal()
{
memset(dp, -1, sizeof(dp));
for(int i = 0; i <= scc; ++i) {
G[i].clear();
}
for(int u = 1; u <= N; ++u) {
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
int a = bel[u], b = bel[v];
if(a != b) {
G[a].push_back(b);
}
}
}
}
int dfs(int u)
{
if(dp[u] != -1) return dp[u];
int res = 0;
for(int i = 0; i < G[u].size(); ++i) {
int v = G[u][i];
res = max(res, dfs(v));
}
dp[u] = res + num[u];
return dp[u];
}
int main()
{
//freopen("in", "r", stdin);
while(~scanf("%d %d", &N, &M)) {
init();
for(int i = 0; i < M; ++i) {
int u, v;
scanf("%d %d", &u, &v);
addedge(u, v);
}
Scc();
deal();
int ans = 0;
for(int i = 1; i <= scc; ++i) {
ans = max(ans, dfs(i));
}
printf("%d\n", ans);
}
return 0;
}
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5303
题意:
N个数,M对数之间的关系,。没有关系的数才能放在同一个集合,求出最少的集合数。
思路:
题目转化为求有向图中节点数最大的节点数集,使得节点数集中任意两个点u和v满足,u可以到达v或者是v可以到达u。(因为互相有关系的点必须放在不同的集合中)
将数之间的关系建有向图,跑 强连通。缩点 得到强连通图。
在强连通图上 dp,求出最长的连续序列。
做过类似的题目:http://blog.csdn.net/sotifish/article/details/44681225
AC.
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
using namespace std;
const int MAXN = 1e5+100;
const int MAXM = 3e5+100;
int N, M;
struct Edge {
int to, next;
}edge[MAXM];
int head[MAXN], tot;
int low[MAXN], dfn[MAXN], stak[MAXN], bel[MAXN], num[MAXN];
int idx, top, scc;
bool instk[MAXN];
void init()
{
tot = 0;
memset(head, -1, sizeof(head));
}
void addedge(int u, int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
head[u] = tot++;
}
void tarjan(int u)
{
int v;
low[u] = dfn[u] = ++idx;
stak[top++] = u;
instk[u] = true;
for(int i = head[u]; ~i; i = edge[i].next) {
v = edge[i].to;
if(!dfn[v]) {
tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(instk[v]) {
low[u] = min(low[u], dfn[v]);
}
}
if(low[u] == dfn[u]) {
scc++;
do {
v = stak[--top];
instk[v] = false;
bel[v] = scc;
num[scc]++;
}while(v!=u);
}
}
void Scc()
{
memset(dfn, 0, sizeof(dfn));
memset(instk, 0, sizeof(instk));
memset(num, 0, sizeof(num));
idx = scc = top = 0;
for(int i = 1; i <= N; ++i) {
if(!dfn[i]) {
tarjan(i);
}
}
}
vector<int> G[MAXN];
int dp[MAXN];
void deal()
{
memset(dp, -1, sizeof(dp));
for(int i = 0; i <= scc; ++i) {
G[i].clear();
}
for(int u = 1; u <= N; ++u) {
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
int a = bel[u], b = bel[v];
if(a != b) {
G[a].push_back(b);
}
}
}
}
int dfs(int u)
{
if(dp[u] != -1) return dp[u];
int res = 0;
for(int i = 0; i < G[u].size(); ++i) {
int v = G[u][i];
res = max(res, dfs(v));
}
dp[u] = res + num[u];
return dp[u];
}
int main()
{
//freopen("in", "r", stdin);
while(~scanf("%d %d", &N, &M)) {
init();
for(int i = 0; i < M; ++i) {
int u, v;
scanf("%d %d", &u, &v);
addedge(u, v);
}
Scc();
deal();
int ans = 0;
for(int i = 1; i <= scc; ++i) {
ans = max(ans, dfs(i));
}
printf("%d\n", ans);
}
return 0;
}
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