CF1366F-Jog Around The Graph
题目:
题目描述:
You are given a simple weighted connected undirected graph, consisting of $ n $ vertices and $ m $ edges.
A path in the graph of length $ k $ is a sequence of $ k+1 $ vertices $ v_1, v_2, \dots, v_{k+1} $ such that for each $ i $ $ (1 \le i \le k) $ the edge $ (v_i, v_{i+1}) $ is present in the graph. A path from some vertex $ v $ also has vertex $ v_1=v $ . Note that edges and vertices are allowed to be included in the path multiple times.
The weight of the path is the total weight of edges in it.
For each $ i $ from $ 1 $ to $ q $ consider a path from vertex $ 1 $ of length $ i $ of the maximum weight. What is the sum of weights of these $ q $ paths?
Answer can be quite large, so print it modulo $ 10^9+7 $ .
输入格式:
The first line contains a three integers $ n $ , $ m $ , $ q $ ( $ 2 \le n \le 2000 $ ; $ n - 1 \le m \le 2000 $ ; $ m \le q \le 10^9 $ ) — the number of vertices in the graph, the number of edges in the graph and the number of lengths that should be included in the answer.
Each of the next $ m $ lines contains a description of an edge: three integers $ v $ , $ u $ , $ w $ ( $ 1 \le v, u \le n $ ; $ 1 \le w \le 10^6 $ ) — two vertices $ v $ and $ u $ are connected by an undirected edge with weight $ w $ . The graph contains no loops and no multiple edges. It is guaranteed that the given edges form a connected graph.
输出格式:
Print a single integer — the sum of the weights of the paths from vertex $ 1 $ of maximum weights of lengths $ 1, 2, \dots, q $ modulo $ 10^9+7 $ .
样例:
样例输入 1:
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| 7 8 25
1 2 1
2 3 10
3 4 2
1 5 2
5 6 7
6 4 15
5 3 1
1 7 3
|
样例输出 1:
样例输入 2:
样例输出 2:
样例输入 3:
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| 15 15 23
13 10 12
11 14 12
2 15 5
4 10 8
10 2 4
10 7 5
3 10 1
5 6 11
1 13 8
9 15 4
4 2 9
11 15 1
11 12 14
10 8 12
3 6 11
|
样例输出 3:
样例输入 4:
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| 5 10 10000000
2 4 798
1 5 824
5 2 558
4 1 288
3 4 1890
3 1 134
2 3 1485
4 5 284
3 5 1025
1 2 649
|
样例输出 4:
思路:
1.先考虑 $k\leq m$ 的情况，可以直接用 $O(N\times M)$ 的暴力 dp 解决
2.$k > M$ 时，最长路径的最后一段一定是在一条边上来回走。 考虑枚举最后来回走的那一条边，再枚举走到那条边的时间，时间
一定在 $M$ 以内。斜率优化 dp 即可
实现:
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| #include "ybwhead/ios.h"
#define int long long
int n, m;
long long q;
const int maxn = 2e3 + 10;
struct edge
{
int v, nxt, w;
} e[maxn << 1];
int head[maxn], tot;
int hv[maxn];
const int mod = 1e9 + 7;
const int inv2 = mod / 2 + 1;
void __ADD(int u, int v, int w)
{
e[++tot].v = v;
e[tot].w = w;
e[tot].nxt = head[u];
head[u] = tot;
hv[v] = max(hv[v], w);
}
void add(int u, int v, int w)
{
__ADD(u, v, w);
__ADD(v, u, w);
}
int d[maxn], nd[maxn];
struct frac
{
long long x, y;
frac(long long a, long long b)
{
if (b < 0)
a = -a, b = -b;
x = a, y = b;
}
bool operator<(frac b)
{
return x * b.y <= y * b.x;
}
};
struct lin
{
long long m, c;
frac xx(const lin &l) { return frac(c - l.c, l.m - m); }
} fi[maxn];
deque<lin> ch;
int cmp(lin a, lin b)
{
if (a.m ^ b.m)
return a.m < b.m;
return a.c > b.c;
}
int cmp1(lin a, lin b)
{
return a.m == b.m;
}
int add(int a, int b)
{
a += b;
if (a >= mod)
a -= mod;
if (a < 0)
a += mod;
return a;
}
int mul(int a, int b)
{
return a * 1ll * b % mod;
}
int calc(int a1, int d, int n)
{
return mul(mul(n, inv2), add(mul(2, a1), mul(add(n, -1), d)));
}
signed main()
{
yin >> n >> m >> q;
for (int i = 1; i <= m; i++)
{
int u, v, w;
yin >> u >> v >> w;
add(u, v, w);
}
memset(d, -0x7f7f7f7f, sizeof(d));
d[1] = 0;
long long ans = 0;
for (int x = 1; x <= m; x++)
{
long long mx = 0;
for (int i = 1; i <= n; i++)
mx = max(mx, (long long)d[i]);
if (x != 1)
ans = add(ans, mx % mod);
for (int i = 1; i <= n; i++)
nd[i] = d[i];
for (int u = 1; u <= n; u++)
{
for (int i = head[u]; i; i = e[i].nxt)
{
int v = e[i].v;
nd[v] = max(nd[v], d[u] + e[i].w);
}
}
for (int i = 1; i <= n; i++)
d[i] = nd[i];
}
for (int i = 1; i <= n; i++)
fi[i] = (lin){hv[i], d[i]};
sort(fi + 1, fi + n + 1, cmp);
int mm = unique(fi + 1, fi + n + 1, cmp1) - fi - 1;
for (int i = 1; i <= mm; i++)
{
// cout << fi[i].m << ' ' << fi[i].c << endl;
while (ch.size() >= 2 && fi[i].xx(ch.back()) < ch.back().xx(ch[ch.size() - 2]))
ch.pop_back();
ch.push_back(fi[i]);
}
long long prv = 0;
q -= m;
for (int i = 0; i < ch.size() - 1; i++)
{
frac f = ch[i].xx(ch[i + 1]);
if (f.x < 0)
continue;
long long lst = min((long long)q, f.x / f.y);
if (lst < prv)
continue;
ans = add(ans, calc((ch[i].c + ch[i].m * prv) % mod, ch[i].m % mod, lst - prv + 1));
prv = lst + 1;
}
ans = add(ans, calc((ch.back().c + ch.back().m * prv) % mod, ch.back().m % mod, q - prv + 1)) % mod;
yout << (ans + mod) % mod << endl;
return 0;
}
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