HDU-1688-Sightseeing

来源：互联网发布：matlab 矩阵元素赋值编辑：程序博客网时间：2024/06/05 06:39

ACM模版

描述

题解

第k短路，这里要求的是第1短路（即最短路），第2短路（即次短路），以及路径条数，最后如果最短路和次短路长度差1，则输出两种路径条数之和，否则只输出最短路条数。

第一次做次短路问题，很有趣哦~~~

代码

#include <iostream>#include <cstring>#include <algorithm>#include <queue>#include <vector>#include <cstdio>using namespace std;const int INF = 0x3f3f3f3f;const int MAXN = 1010;struct Edge{    int to, w;};struct node{    int v, dist;    int tag;    bool friend operator < (const node &a, const node &b)    {        if (a.dist != b.dist)        {            return a.dist > b.dist;        }        return a.v > b.v;    }};priority_queue<node> Q;vector<Edge> graph[MAXN];int cnt[MAXN][3];   //  cnt[v][1]存储最短路径数，cnt[v][2]存储次短路径数int vis[MAXN][3];   //  同理，vis存状态int dis[MAXN][3];   //  dis[v][1]存储最短，dis[v][2]存储次短int N, M, A, B, L;void init(){    for (int i = 0; i < MAXN; i++)    {        cnt[i][1] = cnt[i][2] = 0;        vis[i][1] = vis[i][2] = 0;        dis[i][1] = dis[i][2] = INF;    }}void Dijstra(int st, int ed){    dis[st][1] = 0;    cnt[st][1] = 1;    node q, p;    p.dist = 0;    p.tag = 1;    p.v = st;    Q.push(p);    while (!Q.empty())    {        q = Q.top();        Q.pop();        if (vis[q.v][q.tag])        {            continue;        }        vis[q.v][q.tag] = 1;        for (int i = 0; i < graph[q.v].size(); i++)        {            int v = graph[q.v][i].to;            int w = graph[q.v][i].w;            //  找到一条比“最短”更短的路            if (!vis[v][1] && dis[v][1] > q.dist + w)            {                //  把“最短“变为”次短“                if (dis[v][1] != INF)                {                    dis[v][2] = dis[v][1];                    cnt[v][2] = cnt[v][1];                    p.dist = dis[v][2];                    p.tag = 2;                    p.v = v;                    Q.push(p);                }                //  更新”最短“                dis[v][1] = q.dist + w;                cnt[v][1] = cnt[q.v][q.tag];                p.tag = 1;                p.dist = dis[v][1];                p.v = v;                Q.push(p);            }            //  找到一条”最短“，更新”最短条数“            else if (!vis[v][1] && dis[v][1] == q.dist + w)            {                cnt[v][1] += cnt[q.v][q.tag];            }            //  找到一条比”最短“长，比”次短“短的路，更新”次短“            else if (!vis[v][2] && dis[v][2] > q.dist + w)            {                dis[v][2] = q.dist + w;                cnt[v][2] = cnt[q.v][q.tag];                p.tag = 2;                p.dist = dis[v][2];                p.v = v;                Q.push(p);            }            //  找到一条”次短“，更新”次短条数“            else if (!vis[v][2] && dis[v][2] == q.dist + w)            {                cnt[v][2] += cnt[q.v][q.tag];            }        }    }}int main(){    int T;    scanf("%d", &T);    Edge temp;    int st, ed;    while (T--)    {        init();        for (int i = 0; i < MAXN; i++)        {            graph[i].clear();        }        scanf("%d%d", &N, &M);        for (int i = 0; i < M; i++)        {            scanf("%d%d%d", &A, &B, &L);            temp.to = B;            temp.w = L;            graph[A].push_back(temp);        }        scanf("%d%d", &st, &ed);        Dijstra(st, ed);        if (dis[ed][1] + 1 == dis[ed][2])        {            printf("%d\n", cnt[ed][1] + cnt[ed][2]);        }        else        {            printf("%d\n", cnt[ed][1]);        }    }    return 0;}

参考

《第K短路》

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