POJ - 1062 昂贵的聘礼 Dijkstra

           思路：构造最短路模型，抽象出来一个源点，这个源点到第i个点的费用就是price[i]，然后就能抽象出图来，终点是1.

任意两个人之间都有等级限制，就枚举所有最低等级限制，然后将不再区间[min_lev, min_lev+m]区间的点都删除，就能进行Dijkstra算法了。

AC代码

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <utility>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
#define eps 1e-10
#define inf 0x3f3f3f3f
#define PI pair<int, int>
typedef long long LL;
const int maxn = 100+5;
int price[maxn], lev[maxn];
int cost[maxn][maxn], vis[maxn], d[maxn];
int n, m;

int Dijkstra() {
memset(d, inf, sizeof(d));
d[0] = 0;
for(int i = 0; i <= n; ++i) {
int x = 0, tmp = inf;
for(int y = 0; y <= n; ++y) if(!vis[y] && d[y] <= tmp) tmp = d[x=y];
if(vis[x]) continue;
vis[x] = 1;
for(int y = 0; y <= n; ++y) if(!vis[y]) d[y] = min(d[y], d[x] + cost[x][y]);
}
return d[1];
}

int main() {
while(scanf("%d%d", &m, &n) == 2) {
int r, x, c;
memset(cost, inf, sizeof(cost));
for(int i = 1; i <= n; ++i) {
scanf("%d%d%d", &price[i], &lev[i], &r);
cost[0][i] = price[i]; //源点0到各点的费用
for(int j = 0; j < r; ++j) {
scanf("%d%d", &x, &c);
cost[x][i] = min(cost[x][i], c);
}
}
int ans = inf;
for(int i = 1; i <= n; ++i) {
int min_lev = lev[i]; //最低等级
memset(vis, 0, sizeof(vis));
for(int j = 1; j <= n; ++j) {
if(lev[j] < min_lev || lev[j] - min_lev > m) vis[j] = 1;
}
int dis = Dijkstra();
ans = min(ans, dis);
}
printf("%d\n", ans);
}
return 0;
}

如有不当之处欢迎指出！