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poj1062 昂贵的聘礼 (spfa)

2015-01-09 18:06 225 查看
思路:因为交易过的所有人等级之差最大不能超过m,所以枚举每一个宽为m的范围,其中必须包涵1号节点,遇到范围之外的点不考虑,分别求最短路即可。

代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

struct Edge{
int to, v;
Edge(int a, int b) : to(a), v(b) {}
};

int n, m;
int level[105];
vector<Edge> edges[105];
int dist[105];
int vis[105];
void spfa(int l, int r) {
queue<int> q;
memset(dist, 0x3f, sizeof(dist));
memset(vis, 0, sizeof(vis));
dist[0] = 0;
q.push(0);
vis[0] = 1;
while (!q.empty()) {
int cur = q.front();
q.pop();
for (int i = 0; i < edges[cur].size(); i++) {
Edge e = edges[cur][i];
if (level[e.to] < l || level[e.to] > r) continue;// 限制之外的点不考虑
if (dist[cur] + e.v < dist[e.to]) {
dist[e.to] = dist[cur] + e.v;
if (!vis[e.to]) {
q.push(e.to);
vis[e.to] = 1;
}
}
}
vis[cur] = 0;
}
}

int main() {
while (~scanf("%d %d", &m, &n)) {
int a, b, k;
for (int i = 0; i <= n; i++)
edges[i].clear();
for (int i = 1; i <= n; i++) {
scanf("%d %d %d", &a, &b, &k);
edges[0].push_back(Edge(i, a));
level[i] = b;
for (int j = 1; j <= k; j++) {
scanf("%d %d", &a, &b);
edges[a].push_back(Edge(i, b));
}
}
int ans = 0x3f3f3f3f;
for (int i = level[1] - m; i <= level[1]; i++) {// 枚举等级限制
spfa(i, i + m);
ans = min(ans, dist[1]);
}
cout << ans << endl;
}
return 0;
}
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