CCF-CSP题解 201609-4 交通规划

最小最短路径树。

\(dis[j]==dis[i]+w(i,j)\)时，从\(w(i,j')\)和\(w(i,j)\)考虑。——从0分到100分。

#include <bits/stdc++.h>
const int maxn = 10000;
const int maxm = 100000;

using namespace std;

int to[maxm * 2 + 10];
int w[maxm * 2 + 10];
int nex[maxm * 2 + 10];
int head[maxn + 10], cnt = 0;

void addEdge(int a, int b, int c)
{
    to[cnt] = b; w[cnt] = c;
    nex[cnt] = head[a]; head[a] = cnt++;
    to[cnt] = a; w[cnt] = c;
    nex[cnt] = head[b]; head[b] = cnt++;
}

int edgeVis[maxm + 10];
int dotVis[maxn + 10];
int dis[maxn + 10];
int edgePre[maxn + 10];

void spfa()
{
    memset(edgeVis, 0, sizeof(edgeVis));
    memset(dotVis, 0, sizeof(dotVis));
    memset(dis, 0x3f, sizeof(dis));
    memset(edgePre, -1, sizeof(edgePre));
    queue<int> q;
    dis[1] = 0;
    q.push(1);
    dotVis[1] = 1;
    while (!q.empty())
    {
        int x = q.front(); q.pop();
        dotVis[x] = 0;
        for (int i = head[x]; i != -1; i = nex[i])
        {
            int l = to[i];
            if (dis[l] > dis[x] + w[i])
            {
                dis[l] = dis[x] + w[i];
                if (edgePre[l] != -1)
                    edgeVis[edgePre[l]] = 0;
                edgePre[l] = i / 2;
                edgeVis[edgePre[l]] = 1;
                if (!dotVis[l])
                {
                    q.push(l);
                    dotVis[l] = 1;
                }
            }
            else if (dis[l] == dis[x] + w[i])
            {
                if (w[edgePre[l] * 2] > w[i])
                {
                    edgeVis[edgePre[l]] = 0;
                    edgePre[l] = i / 2;
                    edgeVis[edgePre[l]] = 1;
                }
            }
        }
    }
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);

    memset(head, -1, sizeof(head));
    for (int i = 1, a, b, c; i <= m; i++)
    {
        scanf("%d%d%d", &a, &b, &c);
        addEdge(a, b, c);
    }

    spfa();

    int ans = 0;
    for (int i = 0; i <= cnt - 1; i += 2)
    {
        if (edgeVis[i / 2])
        {
            ans += w[i];
        }
    }

    printf("%d\n", ans);

    return 0;
}