POJ 3463 Sightseeing (第k短路）

~~题目链接~~~

题目大意：给出一个旅游路线，现在要求求它的最短路和比最短路长1的路共有多少条。

思路：看代码吧

#include <cstdio>
#include <cstring>
#define inf 1234567890
using namespace std;
struct node
{
    int v, w, next;
}edge[10002];
struct node2
{
    int cnt,sum;
}dist[1002][4];
int n = 0, m = 0, cnt = 0, s = 0, f = 0, sum = 0, us[1002], head[1002], used[1002];
void swap(node2 &a, node2 &b)
{
    node2 t = a;
    a = b;
    b = t;
}
void dijkstar()
{
    int i = 0, j = 0, p = 0, v = 0, w = 0, k = 0, flag = 0, min = 0;
    dist[0][0].sum = inf;
    dist[s][0].sum = 0;
    dist[s][0].cnt = 1;
    while(1)
    {
        k = 0;
        min = inf;
        for(i = 1; i<=n; i++)//应该先更新完最短路在的条数在更新次短路， 每回选出当前更新状态的最短距离
        {
            if(us[i] == 0 && dist[i][0].sum<min)
            {
                k = i;
                min = dist[i][0].sum;
            }
            else if(us[i] == 1 && dist[i][1].sum<min)
            {
                min = dist[i][1].sum;
                k = i;
            }
        }
        if(k == 0) return ;//表示当前已没有路， 必须要有这个判断， 不然超时
        if(k == f && us[k] == 1) return ;//只要一满足这个条件， 终点的最短路与此短路必定统计完了，此时结束
        for(p = head[k]; p != -1; p = edge[p].next)
        {
            v = edge[p].v;
            w = edge[p].w;
            if(us[v]<3 && dist[k][us[k]].sum+w < dist[v][2].sum)//小于第三短路
            {
                dist[v][2].sum = dist[k][us[k]].sum+w;
                dist[v][2].cnt = dist[k][us[k]].cnt;
                flag = 0;
                for(j = 0; j<2; j++)//看这点是否出现过
                {
                    if(dist[v][j].sum==dist[v][2].sum)
                    {
                        flag = 1;
                        dist[v][j].cnt += dist[v][2].cnt;
                        dist[v][2].sum = inf;
                        dist[v][2].cnt = 0;
                        break;
                    }
                }
                if(!flag)//如果没有出现过，把他更新到正确距离
                {
                    for(j = 2; j>0; j--)//调整最短路， 次短路
                        if(dist[v][j].sum<dist[v][j-1].sum)
                            swap(dist[v][j], dist[v][j-1]);
                }
            }
        }
        us[k]++;
    }
}
int main()
{
    int i = 0, j = 0, t = 0, u = 0, v = 0, w = 0;
    scanf("%d", &t);
    while(t--)
    {
        cnt = 0;
        for(i = 0; i<1002; i++)
        {
            head[i] = -1;
            us[i] = 0;
            for(j = 0; j<4; j++)
            {
                dist[i][j].sum = inf;
                dist[i][j].cnt = 0;
            }
        }
        scanf("%d %d", &n, &m);
        for(i = 0; i<m; i++)
        {
            scanf("%d %d %d", &u, &v, &w);
            edge[cnt].v = v;
            edge[cnt].next = head[u];
            edge[cnt].w = w;
            head[u] = cnt++;
        }
        scanf("%d %d", &s, &f);
        dijkstar();
        if(dist[f][1].sum-dist[f][0].sum == 1)
            printf("%d\n", dist[f][1].cnt+dist[f][0].cnt);
        else
            printf("%d\n", dist[f][0].cnt);
    }
    return 0;
}