POJ - 1797 Heavy Transportation 单源最短路

思路：d(i)表示到达节点i的最大能运输的重量，转移方程d(i) = min(d(u), limit(u, i));注意优先队列应该以重量降序排序来重载小于符号。

AC代码

#include <cstdio>
#include <cmath>
#include <cctype>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 1000 + 5;
int d[maxn];
bool vis[maxn];
struct Edge{
	int from, to, dist;
	Edge(){}
	Edge(int u, int v, int d):from(u), to(v), dist(d) {}
};
vector<Edge>edge;
struct HeapNode{
	int d, u;
	HeapNode() {}
	HeapNode(int d, int u):d(d), u(u) {}
	bool operator < (const HeapNode &p) const {
		return d < p.d;
	}
};
vector<int>G[maxn];

void init(int n) {
	edge.clear();
	for(int i = 0; i <= n; ++i) G[i].clear();
}

void addEdge(int from, int to, int dist) {
	edge.push_back(Edge(from, to, dist));
	int m = edge.size();
	G[from].push_back(m-1);
}

void dijkstra(int s) {
	priority_queue<HeapNode>q;
	memset(d, 0, sizeof(d));
	d[s] = inf;
	memset(vis, 0, sizeof(vis));
	q.push(HeapNode(inf, 1));
	while(!q.empty()) {
		HeapNode p = q.top(); q.pop();
		int u = p.u;
		if(vis[u]) continue;
		for(int i = 0; i < G[u].size(); ++i) {
			Edge &e = edge[G[u][i]];
			int w = min(d[u], e.dist);
			if(d[e.to] < w) {
				d[e.to] = w;
				q.push(HeapNode(d[e.to], e.to));
			}
		}
	}
}

int main() {
	int T, n, m;
	scanf("%d", &T);
	int kase = 1;
	while(T--) {
		scanf("%d%d", &n, &m);
		init(n);
		int u, v, dis;
		for(int i = 0; i < m; ++i) {
			scanf("%d%d%d", &u, &v, &dis);
			addEdge(u, v, dis);
			addEdge(v, u, dis);
		}
		dijkstra(1);
		printf("Scenario #%d:\n", kase++);
		printf("%d\n\n", d[n]);
	}
	return 0;
}

如有不当之处欢迎指出！