ACM模版

描述

题解

字典树+最短路即可，糟糕的题目，表达有些问题，导致理解错了N……，还是我太渣了。

一开始以为需要搞个N*N的矩阵（代码One），会爆内存，然后就用了堆优化（邻接表，代码Two），可是因为没有正确理解N的表示，还是超时了，改了改对了。

代码

One：

#include <iostream>
#include <cstring>
#include <map>

using namespace std;

/* * 单源最短路径，Dijkstra算法，邻接矩阵形式，复杂度为O(n^2) * 求出源beg到所有点的最短路径，传入图的顶点数和邻接矩阵cost[][] * 返回各点的最短路径lowcost[]，路径pre[]，pre[i]记录beg到i路径上的父节点，pre[beg] = -1 * 可更改路径权类型，但是权值必须为非负，下标0~n-1 */
const int MAXN = 160;
const int INF = 0x3f3f3f3f; // 表示无穷
const int MAXS = 31;
bool vis[MAXN];
int pre[MAXN];

void Dijkstra(int cost[][MAXN], int lowcost[], int n, int beg)
{
    for (int i = 0; i < n; i++)
    {
        lowcost[i] = INF;
        vis[i] = false;
        pre[i] = -1;
    }
    lowcost[beg] = 0;
    for (int j = 0; j < n; j++)
    {
        int k = -1;
        int min = INF;
        for (int i = 0; i < n; i++)
        {
            if (!vis[i] && lowcost[i] < min)
            {
                min = lowcost[i];
                k = i;
            }
        }
        if (k == -1)
        {
            break;
        }
        vis[k] = true;
        for (int i = 0; i < n; i++)
        {
            if (!vis[i] && lowcost[k] + cost[k][i] < lowcost[i])
            {
                lowcost[i] = lowcost[k] + cost[k][i];
                pre[i] = k;
            }
        }
    }
}

map<string, int> m;
map<string, int>::iterator it;
int cost[MAXN][MAXN];
int lowcost[MAXN];

int main(int argc, const char * argv[])
{
    int N;
    char start[MAXS], end[MAXS];
    char s[MAXS], e[MAXS];
    int t;

    while (cin >> N, N != -1)
    {
        m.clear();
        memset(cost, 0x3f, sizeof(cost));

        scanf("%s %s", start, end);
        m[start] = 0;
        m[end] = 1;

        int pos = 2;
        for (int i = 0; i < N; i++)
        {
            scanf("%s %s %d", s, e, &t);
            it = m.find(s);
            if (it == m.end())
            {
                m[s] = pos++;
            }
            it = m.find(e);
            if (it == m.end())
            {
                m[e] = pos++;
            }
            cost[m[s]][m[e]] = cost[m[e]][m[s]] = t;
        }

        if (m[start])
        {
            printf("0\n");
            continue;
        }

        Dijkstra(cost, lowcost, pos, 0);

        if (lowcost[1] == INF)
        {
            printf("-1\n");
        }
        else
        {
            printf("%d\n", lowcost[1]);
        }
    }

    return 0;
}

Two：

#include <iostream>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <map>

using namespace std;

/* * 使用优先队列优化Dijkstra算法 * 复杂度O(ElongE) * 注意对vector<Edge> E[MAXN]进行初始化后加边 */

const int INF = 0x3f3f3f3f;
const int MAXN = 10010;

struct qNode
{
    int v;
    int c;
    qNode(int _v = 0, int _c = 0) : v(_v), c(_c) {}
    bool operator < (const qNode &r)const
    {
        return c > r.c;
    }
};

struct Edge
{
    int v;
    int cost;
    Edge(int _v = 0, int _cost = 0) : v(_v), cost(_cost) {}
};

vector<Edge> E[MAXN];
bool vis[MAXN];
int dist[MAXN];

void Dijkstra(int n, int start)     // 点的编号从1开始
{
    memset(vis, false, sizeof(vis));
    memset(dist, 0x3f, sizeof(dist));
    priority_queue<qNode> que;

    while (!que.empty())
    {
        que.pop();
    }
    dist[start] = 0;
    que.push(qNode(start, 0));
    qNode tmp;

    while (!que.empty())
    {
        tmp = que.top();
        que.pop();
        int u = tmp.v;
        if (vis[u])
        {
            continue;
        }
        vis[u] = true;
        for (int i = 0; i < E[u].size(); i++)
        {
            int v = E[tmp.v][i].v;
            int cost = E[u][i].cost;
            if (!vis[v] && dist[v] > dist[u] + cost)
            {
                dist[v] = dist[u] +cost;
                que.push(qNode(v, dist[v]));
            }
        }
    }
}

void addEdge(int u, int v, int w)
{
    E[u].push_back(Edge(v, w));
}

const int MAXS = 31;

map<string, int> m;
map<string, int>::iterator it;

int main(int argc, const char * argv[])
{
    int N;
    char start[MAXS], end[MAXS];
    char s[MAXS], e[MAXS];
    int t;

    while (cin >> N, N != -1)
    {
        m.clear();
        for (int i = 0; i <= N; i++)
        {
            E[i].clear();
        }

        scanf("%s %s", start, end);
        m[start] = 0;
        m[end] = 1;

        int pos = 2;
        for (int i = 0; i < N; i++)
        {
            scanf("%s %s %d", s, e, &t);
            it = m.find(s);
            if (it == m.end())
            {
                m[s] = pos++;
            }
            it = m.find(e);
            if (it == m.end())
            {
                m[e] = pos++;
            }
            addEdge(m[s], m[e], t);
            addEdge(m[e], m[s], t);
        }

        Dijkstra(pos, 0);

        if (dist[1] == INF)
        {
            printf("-1\n");
        }
        else
        {
            printf("%d\n", dist[1]);
        }
    }

    return 0;
}

参考

《最短路》