POJ1062 昂贵的聘礼

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;
}