[ABC271E] Subsequence Path 题解

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收起

[ABC271E] Subsequence Path 题解

思路解析

很好的一道题，很有迷惑性，表面上是一道图论实际上是 dp，定义 $f_i$ 为从 $1$ 到 $i$ 的最短 “好路”。先把每条边对应的起点，终点和边权记录下来，然后输入每个 $e$，由于是子序列顺序不会改变，因此可以顺序进行操作。对于每一个 $e$，都有选和不选两种情况，若选，则相当于最短路算法中的松弛操作，转移 $f_{a_e}\leftarrow min(f_{a_e},f_{b_e}+c_e)$（$a_i,b_i,c_i$ 分别代表第 $i$ 条边的起点，终点和边权）。

code

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 2e5 + 10;
ll n, m, k;
ll a[N], b[N], c[N];
ll e;
ll f[N];
int main() {
	cin >> n >> m >> k;
	for(int i = 1; i <= m; i++) {
		cin >> a[i] >> b[i] >> c[i];
	}
	memset(f, 0x3f, sizeof(f));
	f[1] = 0;
	for(int i = 1; i <= k; i++) {
		cin >> e;
		f[b[e]] = min(f[b[e]], f[a[e]] + c[e]);
	}
	if(f[n] < 1e18) {
		cout << f[n];
	}
	else {
		cout << -1;
	}
	return 0;
}