POJ1062 昂贵的聘礼(dijkstra)

题目链接：点击打开链接

读入数据后，扫一遍物品，每一遍都根据等级关系标记这个物品是否可以使用，进行一次dijkstra，求出最小花费金币数。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 105;
int m, n, price[MAXN][MAXN], lv[MAXN], dis[MAXN], x[MAXN];
bool vis[MAXN];
int dijkstra()
{
	for(int i = 1; i <= n; ++i)
		dis[i] = price[0][i];
	int x, y;
	for(int i = 1; i <= n; ++i) {
		x = 0, y = INF;
		for(int j = 1; j <= n; ++j)
			if(!vis[j] && y > dis[j]) {
				y = dis[j];
				x = j;
			}
		if(!x) break;
		vis[x] = true;
		for(int j = 1; j <= n; ++j)
			if(!vis[j] && price[x][j] > 0 && dis[j] > dis[x] + price[x][j])
				dis[j] = dis[x] + price[x][j];
	}
	return dis[1];
}
int main(int argc, char const *argv[])
{
	while(scanf("%d%d", &m, &n) != EOF) {
		memset(price, 0, sizeof(price));
		memset(lv, 0, sizeof(lv));
		memset(dis, 0, sizeof(dis));
		memset(x, 0, sizeof(x));
		memset(vis, false, sizeof(vis));
		for(int i = 1; i <= n; ++i) {
			scanf("%d%d%d", &price[0][i], &lv[i], &x[i]);
			for(int j = 1; j <= x[i]; ++j) {
				int t, u;
				scanf("%d%d", &t, &u);
				price[t][i] = u;
			}
		}
		int tmp, ans, max_lv;
		for(int i = 1; i <= n; ++i) {
			max_lv = lv[i];
			for(int j = 1; j <= n; ++j)
				if(lv[j] > max_lv || max_lv - lv[j] > m) vis[j] = true;
				else vis[j] = false;
			tmp = dijkstra();
			ans = min(ans, tmp);
		}
		printf("%d\n", ans);
	}
	return 0;
}