POJ1062《昂贵的聘礼》方法：dijkstra

娶酋长媳妇需要10000，但是在拥有物品2时就需要8000，有物品3时需要5000，并且交易时不能和超过等级差的人先后交易，所以每一次先以当前等级为最大等级并且不超过等级差，dis[1]为目标点，运用dijkstra求单源最短路径，每一次的源点是从最小的dis[i]开始。

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;const int maxn = 16843009;int M, N; //M等级差距限制，N物品总数int price[101][101]; //在a为替代品的前提下的b优惠价格，price[0][i]表示原价格int lv[101]; //主人等级int replace[101]; //替代品数int dis[101];bool vist[101];int dijkstra(){for (int i = 1; i <= N; ++i)dis[i] = price[0][i];for (int i = 1; i <= N; ++i) {int node = 0, sp = maxn; //未访问点的编号，和最小距离for (int j = 1; j <= N; ++j) {if (!vist[j] && dis[j] < sp) {node = j;sp = dis[j];}}if (node == 0) //无法找到未访问点的编号，跳出break;vist[node] = true; //加入已访问点for (int j = 1; j <= N; ++j) if (!vist[j] && dis[node] + price[node][j] < dis[j])dis[j] = dis[node] + price[node][j];}return dis[1];}int main(){//freopen("temp.txt", "r", stdin);memset(price, 1, sizeof(price));// 0x01010101为16843009memset(dis, 1, sizeof(dis));memset(vist, 0, sizeof(vist)); // falsecin >> M >> N;int T, V; //T替代品编号，V优惠价格for (int i = 1; i <= N; ++i) {cin >> price[0][i] >> lv[i] >> replace[i];for (int j = 1; j <= replace[i]; ++j) {cin >> T >> V;price[T][i] = V;}}int maxlv, min_price = maxn; //最大阶级for (int i = 1; i <= N; ++i) {maxlv = lv[i]; //把当前阶级看作最大阶级for (int j = 1; j <= N; ++j) {if (lv[j] > maxlv || maxlv - lv[j] > M) { //先剔除比当前阶级大的，和超出阶级差的vist[j] = true;} else {vist[j] = false;}}int temp_price = dijkstra();if (temp_price < min_price)min_price = temp_price;}cout << min_price << endl;return 0;}