CH6101/洛谷P1073 最优贸易（最短路）

男娘i 2021-11-17 04:04 228阅读 0赞

## 题意：从图上找到一条从1到n的路径，使路径上能选出两个点p，q（先经过p后经过q），并且“节点q的权值减去节点p的权值”最大。 ##

## 分析：因为要考虑两个点先后顺序，可以采用建反图的方法。详见《算法竞赛进阶指南》P357。 ##

## 代码： ##

#include<bits/stdc++.h>
    using namespace std;
    const int N = 1e5 + 6, M = 1e6 + 6;
    int n, m, tot, Price[N], Ans;
    int Head[N], Side[M], Next[M], ans[N];
    int fHead[N], fSide[M], fNext[M], fans[N];
    priority_queue<pair<int, int> > q;
    priority_queue<pair<int, int> > fq;
    
    int main() {
    	cin >> n >> m;
    	for (int i = 1; i <= n; i++) scanf("%d", &Price[i]);
    	for (int i = 1; i <= m; i++) {
    		int x, y, z;
    		scanf("%d %d %d", &x, &y, &z);
    		Side[++tot] = y;
    		Next[tot] = Head[x];
    		Head[x] = tot;
    		fSide[tot] = x;
    		fNext[tot] = fHead[y];
    		fHead[y] = tot;
    		if (z == 2) {
    			Side[++tot] = x;
    			Next[tot] = Head[y];
    			Head[y] = tot;
    			fSide[tot] = y;
    			fNext[tot] = fHead[x];
    			fHead[x] = tot;
    		}
    	}
    	memset(ans, 0x3f, sizeof(ans));
    	memset(fans, 0xcf, sizeof(fans));
    	ans[1] = Price[1];
    	fans[n] = Price[n];
    	q.push(make_pair(-ans[1], 1));
    	fq.push(make_pair(fans[n], n));
    	while (q.size()) {
    		int x = q.top().second;
    		q.pop();
    		for (int i = Head[x]; i; i = Next[i]) {
    			int y = Side[i];
    			if (ans[y] > ans[x]) {
    				ans[y] = ans[x];
    				ans[y] = min(ans[y], Price[y]);
    				q.push(make_pair(-ans[y], y));
    			}
    		}
    	}
    	while (fq.size()) {
    		int x = fq.top().second;
    		fq.pop();
    		for (int i = fHead[x]; i; i = fNext[i]) {
    			int y = fSide[i];
    			if (fans[y] < fans[x]) {
    				fans[y] = fans[x];
    				fans[y] = max(fans[y], Price[y]);
    				fq.push(make_pair(fans[y], y));
    			}
    		}
    	}
    	for (int i = 1; i <= n; i++) Ans = max(Ans, fans[i]-ans[i]);
    	cout << Ans << endl;
    	return 0;
    }