POJ - 1062 昂贵的聘礼 （最短路变形）

中文题

解题思路：将酋长最为源点，用一个结构体纪录每个点到酋长的最短路，和级别范围，在更新新节点时，就可以根据级别范围判断能否更新了

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#define N 110
#define M 10010
#define INF 0x3f3f3f3f
using namespace std;

struct People {
    int price, level, x;
}P[N];

struct change{
    int no, price;
}C[M];

struct state {
    int dis, high, low;
}S[N];

int n, m, cnt;
int dis[N][N];
bool vis[N];

void init() {

    cnt = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d%d%d", &P[i].price, &P[i].level, &P[i].x);
        for (int j = 0; j < P[i].x; j++) {
            scanf("%d%d", &C[cnt].no, &C[cnt].price);
            cnt++;
        }
    }

    memset(dis, 0x3f, sizeof(dis));
    cnt = 0;

    for (int i = 1; i <= n; i++) {
        dis[0][i] = P[i].price;
        for (int j = 0; j < P[i].x; j++) {
            if (abs(P[i].level - P[C[cnt].no].level) <= m) {
                dis[i][C[cnt].no] = C[cnt].price;
            }
            cnt++;
        }
    }
}

void dijkstra() {
    memset(vis, 0, sizeof(vis));
    for (int i = 2; i <= n; i++) {
        S[i].dis = INF;
        S[i].high = P[i].level;
        S[i].low = P[i].level;
    }

    S[1].dis = 0;
    S[1].high = P[1].level;
    S[1].low = P[1].level;

    for (int i = 1; i < n; i++) {
        int x, t = INF;
        for (int j = 1; j <= n; j++) {
            if (!vis[j] &&  S[j].dis < t) {
                t = S[j].dis;
                x = j;
            }
        }
        vis[x] = 1;

        for (int j = 1; j <= n; j++) {
            if (!vis[j] && abs(S[x].high - S[j].low) <= m && abs(S[x].low - S[j].high) <= m && S[x].dis + dis[x][j] <= S[j].dis) {
                S[j].dis = S[x].dis + dis[x][j];
                S[j].low = min(S[j].low, S[x].low);
                S[j].high = max(S[j].high, S[x].high);
            }
        }
    }

    int ans = INF;
    for (int i = 1; i <= n; i++) {
        ans = min(ans, S[i].dis + dis[0][i]);
    }
    printf("%d\n", ans);

}

int main() {
    scanf("%d%d", &m, &n);
    init();
    dijkstra();
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。