POJ1062 昂贵的聘礼

来源：互联网发布：js购物车思路编辑：程序博客网时间：2024/05/19 00:50

一.题目链接：http://poj.org/problem?id=1062

二.思路：设起点为0，枚举每个区间，把超出区间的置为已访问，然后Dijkstra求出0到1的最短路，取所有区间中最短的一个。

三.代码：

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <queue>using namespace std;const int MAX_SIZE = 102,          INF = 1<<30,          MOD = 1000000007;int graph[MAX_SIZE][MAX_SIZE];int rankGap, stuffNum;int Rank[MAX_SIZE];void initVisited(int l, int r, bool visited[]){    int i;    for(i = 1; i <= stuffNum; i++)        if(Rank[i] < l || Rank[i] > r)            visited[i] = true;}int Dijkstra(int l, int r){    bool visited[MAX_SIZE];    int dist[MAX_SIZE], i, j, minPath, startPos;    memset(dist, 1, sizeof(dist));    memset(visited, 0, sizeof(visited));    initVisited(l, r, visited);    visited[0] = true;    dist[0] = 0;    for(i = 0; i <= stuffNum; i++)        dist[i] = graph[0][i];    for(i = 0; i < stuffNum; i++){        minPath = INF;        for(j = 0; j <= stuffNum; j++){            if(minPath > dist[j] && !visited[j]){                minPath = dist[j];                startPos = j;            }        }        visited[startPos] = true;        for(j = 0; j <= stuffNum; j++){            if(dist[j] > minPath + graph[startPos][j] && !visited[j])                dist[j] = minPath + graph[startPos][j];        }    }    return dist[1];}void printGraph(){    for(int i = 0; i <= stuffNum; i++){        for(int j = 0; j <= stuffNum; j++){            if(graph[i][j] == 16843009)                cout<<"INF ";            else                cout<<graph[i][j]<<" ";        }        cout<<endl;    }}int main(){    //freopen("in.txt", "r", stdin);    int i, j, beginPos, weight, pathNum;    memset(graph, 1, sizeof(graph));    cin>>rankGap>>stuffNum;    for(i = 1; i <= stuffNum; i++){        cin>>weight>>Rank[i]>>pathNum;        graph[0][i] = weight;        for(j = 0; j < pathNum; j++){            cin>>beginPos>>weight;            graph[beginPos][i] = weight;        }    }    int res = INF;    for(i = Rank[1] - rankGap; i <= Rank[1]; i++)        res = min(res, Dijkstra(i, i + rankGap));    cout<<res<<endl;}

0 0