九度OJ 1162：I Wanna Go Home（我想回家） （最短路径）

时间限制：1 秒

内存限制：32 兆

特殊判题：否

提交：870

解决：415

题目描述：

The country is facing a terrible civil war----cities in the country are divided into two parts supporting different leaders. As a merchant, Mr. M does not pay attention to politics but he actually knows the severe situation, and your task is to help him
reach home as soon as possible. 

    "For the sake of safety,", said Mr.M, "your route should contain at most 1 road which connects two cities of different camp."

    Would you please tell Mr. M at least how long will it take to reach his sweet home?

输入：

The input contains multiple test cases.

    The first line of each case is an integer N (2<=N<=600), representing the number of cities in the country.

    The second line contains one integer M (0<=M<=10000), which is the number of roads.

    The following M lines are the information of the roads. Each line contains three integers A, B and T, which means the road between city A and city B will cost time T. T is in the range of [1,500].

    Next part contains N integers, which are either 1 or 2. The i-th integer shows the supporting leader of city i. 

    To simplify the problem, we assume that Mr. M starts from city 1 and his target is city 2. City 1 always supports leader 1 while city 2 is at the same side of leader 2. 

    Note that all roads are bidirectional and there is at most 1 road between two cities.

Input is ended with a case of N=0.

输出：

For each test case, output one integer representing the minimum time to reach home.

    If it is impossible to reach home according to Mr. M's demands, output -1 instead.

样例输入：

2
1
1 2 100
1 2
3
3
1 2 100
1 3 40
2 3 50
1 2 1
5
5
3 1 200
5 3 150
2 5 160
4 3 170
4 2 170
1 2 2 2 1
0

样例输出：

100
90
540

来源：

2011年北京大学计算机研究生机试真题

思路：

题目大意是N个城市分属于两个敌对集团，要从城市1到城市2（分别属于两个集团），只能有一条路跨集团，求最短路径。

我的思路是，求城市1到其集团中其它城市的最短路径，城市2也同样，然后对集团间存在的路径i到j，求d(1,i)+d(i,j)+d(j,2)的最小值。

代码：

#include <stdio.h>

#define N 600
#define M 10000
#define INF 1e8

int n;
int D[N][N];
int support[N], visit[2][N], dis[2][N];

void init()
{
    for (int i=0; i<n; i++)
    {
        support[i] = 0;
        visit[0][i] = visit[1][i] = 0;
        dis[0][i] = dis[1][i] = INF;
        for (int j=0; j<n; j++)
        {
            D[i][j] = INF;
        }
    }
}

void printdis(int s)
{
    int i;
    for (i=0; i<n; i++)
    {
        if (support[i] == s)
            printf("%d\n", dis[s][i]);
    }
    printf("\n");
}

void dijkstra(int s)
{
    int i, j;
    for (i=0; i<n; i++)
    {
        if (support[i] == s)
            dis[s][i] = D[s][i];
    }
    dis[s][s] = 0;
    visit[s][s] = 1;
    //printdis(s);

    int mind;
    int k;
    for (i=0; i<n; i++)
    {
        mind = INF;
        for (j=0; j<n; j++)
        {
            if ( support[j] == s && !visit[s][j] && (dis[s][j]<mind) )
            {
                mind = dis[s][j];
                k = j;
            }
        }
        if (mind == INF)
            break;
        visit[s][k] = 1;
        for (j=0; j<n; j++)
        {
            if ( support[j] == s && !visit[s][j] && (dis[s][k]+D[k][j] < dis[s][j]) )
            {
                dis[s][j] = dis[s][k]+D[k][j];
            }
        }
    }
    //printdis(s);
}

int Min(int a, int b)
{
    return (a<b) ? a : b;
}

int main(void)
{
    int m, i, j;
    int a, b, d;
    int min;

    while (scanf("%d", &n) != EOF && n)
    {
        init();
        scanf("%d", &m);
        for(i=0; i<m; i++)
        {
            scanf("%d%d%d", &a, &b, &d);
            D[a-1][b-1] = D[b-1][a-1] = d;
        }
        for(i=0; i<n; i++)
        {
            scanf("%d", &a);
            support[i] = a-1;
        }       

        dijkstra(0);
        dijkstra(1);

        min = INF;
        for (i=0; i<n; i++)
        {
            for (j=0; j<n; j++)
            {
                if (support[i] == 0 && support[j] == 1)
                {
                    min = Min(dis[0][i] + D[i][j] + dis[1][j], min);
                }
            }
        }
        if (min == INF)
            printf("-1\n");
        else
            printf("%d\n", min);
    }

    return 0;
}
/**************************************************************
    Problem: 1162
    User: liangrx06
    Language: C
    Result: Accepted
    Time:20 ms
    Memory:2336 kb
****************************************************************/