标签:roo source mes min bre 重复 int ack max
给定$n$个点$m$条边有向图及边权$w$,第$i$次经过一条边边权为$w-1-2.-..-i$,$w\ge 0$给定起点$s$问从起点出发最多能够得到权和,某条边可重复经过
有向图能够重复经过的边当且仅当成环,所以tarjan缩点成DAG,缩点后每个点内的权值可以通过二分算出,假设最大的$n$使得$w-\frac{n(n+1)}{2}\ge 0$,那么该点值为$(n+1)w-\frac{n(n+1)(n+2)}{6}$,通过对DAG进行dp算出最长路就是答案
#include <bits/stdc++.h>
#define inf 0x7f7f7f7f
using namespace std;
typedef long long LL;
const int N = 1000005;
int n, m, x, y, w, s;
int head[N], nxt[N], to[N], val[N], cnt;
inline void init(){memset(head, -1, sizeof(head)); cnt = 0;}
inline void add(int u, int v, int w) {to[cnt] = v, val[cnt] = w, nxt[cnt] = head[u], head[u] = cnt++;}
int head2[N], nxt2[N], to2[N], cnt2; LL val2[N];
inline void init2() {memset(head2, -1, sizeof(head)); cnt2 = 0;}
inline void add2(int u, int v, LL w) {to2[cnt2] = v, val2[cnt2] = w, nxt2[cnt2] = head2[u], head2[u] = cnt2++;}
int dfs_ind = 1, dfn[N], low[N], sccno[N], scc_cnt = 0;
LL w_[N];
stack<int> st;
void tarjan(int u) {
dfn[u] = low[u] = dfs_ind++;
st.push(u);
for(int i = head[u]; ~i; i = nxt[i]) {
int v = to[i];
if(!dfn[v]) {tarjan(v); low[u] = min(low[u], low[v]);}
else if(!sccno[v]) {low[u] = min(low[u], dfn[v]);}
}
if(dfn[u] == low[u]) {
scc_cnt++;
while(1) {
int x = st.top(); st.pop();
sccno[x] = scc_cnt;
if(x == u) break;
}
}
}
inline LL cal(LL x) {
LL n = sqrt(2.0 * x + 0.25) - 0.5;
return (n + 1) * x - (n + 1) * (n + 2) * n / 6;
}
void DAG() {
memset(dfn,0,sizeof(dfn));
memset(low,0,sizeof(low));
memset(sccno,0,sizeof(sccno));
memset(w_,0,sizeof(w_));
init2();
for(int i = 1; i <= n; ++i) if(!dfn[i]) tarjan(i);
for(int i = 1; i <= n; ++i) {
for(int j = head[i]; ~j; j = nxt[j]) {
int v = to[j];
if(sccno[i] != sccno[v]) {
add2(sccno[i], sccno[v], 1LL * val[j]);
}else w_[sccno[i]] += cal(val[j]);
}
}
}
LL dp[N];
void dfs(int u) {
if(~dp[u]) return;
dp[u] = w_[u];
for(int i = head2[u]; ~i; i = nxt2[i]) {
dfs(to2[i]);
dp[u] = max(dp[u], w_[u] + dp[to2[i]] + val2[i]);
}
}
int main() {
init();
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; ++i) {
scanf("%d%d%d", &x, &y, &w);
add(x, y, w);
}
scanf("%d", &s);
DAG();
memset(dp, -1, sizeof(dp));
dfs(sccno[s]);
cout << dp[sccno[s]] << endl;
return 0;
}
【Codeforces】894E.Ralph and Mushrooms Tarjan缩点+DP
标签:roo source mes min bre 重复 int ack max
原文地址:http://www.cnblogs.com/ogiso-setsuna/p/7930209.html