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最新推荐文章于 2019-05-15 00:01:00 发布

原创最新推荐文章于 2019-05-15 00:01:00 发布 · 370 阅读

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本文介绍了一种结合强连通分量与树形动态规划的算法实现，通过实例展示了如何处理图论问题，特别是如何在有向图中寻找最优路径。文章首先定义了所需的数据结构，并详细解释了DFS遍历和树形DP的过程。

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#include <cstdio>
#include <iostream>
#include <map>
#include <cstring>
#include <queue>
#include <stack>
#include <functional>
#include <algorithm>
#include <cmath>
using namespace std;
#define LL long long
#define mini 1e-5
int n, m;
struct edge{
int u, v, next;
};
struct nedge{
int x;
nedge *ne;
};
int low[20050], dfn[20050], inde = 0, mark[20050];
LL wi[20050], f[20050];
bool has[20050],hass[20050];
edge ed[100050];
nedge *nhead[20050],ned_arr[20050],*ned(ned_arr);
int ncnt;
int head[100050],llll[100050];
stack<int>q;
void dfs(int i){
if (has[i]){

}
has[i] = true;
low[i] = dfn[i] = ++inde;
q.push(i);
hass[i] = true;
for (int j = head[i]; j != -1; j = ed[j].next){
int v = ed[j].v;
if (!has[v]){
dfs(v);
low[i] = min(low[i], low[v]);
}
else if(hass[v]){
low[i] = min(low[i], dfn[v]);
}
}
if (dfn[i] == low[i]){
mark[i] = i;
f[i] += wi[i];
while (q.top() != i){
f[i]+= wi[q.top()];
mark[q.top()] = i;
hass[q.top()] = false;
q.pop();
}
hass[i] = false;
q.pop();
ncnt++;
}
}
LL topu(int a){
LL an = f[a];

for (nedge *i = nhead[a]; i; i = i->ne){
an =max( f[a]+topu(i->x),an);
}
return an;
}
int main()
{
int t1, t2;
ncnt = 0;
cin >> n >> m;
memset(head, -1, sizeof(head));
memset(has, false, sizeof(has));
memset(hass, false, sizeof(has));
memset(low, -1, sizeof(low));
memset(dfn, -1, sizeof(dfn));
memset(f, 0, sizeof(f));
memset(mark, -1, sizeof(mark));
for (int i = 1; i <= n; i++){
cin >> wi[i];
}
int cnt = 0,start;
for (int i = 0; i < m; i++){
cin >> t1 >> t2;
ed[cnt].u = t1;
ed[cnt].v = t2;
ed[cnt].next = head[t1];
head[t1] = cnt;
cnt++;
}
dfs(1);
int lcnt=0;
for (int i = 1; i <= n; i++){
if (mark[i] == -1){
continue;
}
if (dfn[i] == low[i]){
bool flag = false;
for (int j = head[i]; j != -1; j = ed[j].next){
if (mark[ed[j].v]!=mark[i]){
flag = true;
ned->x = mark[i];
ned->ne = nhead[mark[ed[j].v]];
nhead[mark[ed[j].v]] = ned++;
}
}
if (!flag){
llll[lcnt++] = mark[i];
}
}
}
LL ans = 0;
for (int i = 0; i < lcnt; i++){
int k = llll[i];
ans = max(topu(k), ans);
}
cout << ans << endl;
//system("pause");
return 0;
}WA30分，不知为何