UVALive - 4080 Warfare And Logistics (SPFA+最短路树)

题目大意：有N个点，M条路，如果两条路不连通的话，就将这两条路的距离设置为L

现在要求你求出每两点之间的最短距离和

接着要求

求出炸断 给出的M条路中的一条路后，每两点之间的最短距离和的最大值

解题思路：这题跟HDU-2433类似，不过这题的权值是不一样的

但具体的思路是差不多的

先预处理出以每个点为源点的最短路树，并纪录每个点的pre和以每个点为源点的最短距离和，这样就可以求出每两点之间的最短距离和了

接着依次删边，如果删除的边不在该点最短路树上，那么就可以用预处理纪录的以该点为源点的最短距离和了

如果在最短路树上，就记录一下删除的边，再跑一遍以该点为源点的最短路即可

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;

#define N 110
#define M 15010
#define INF 0x3f3f3f3f

struct Edge {
    int to, next, dist;
}E[M];

struct Node {
    int x, y, id;
}node[M];

int n, m, l, tot;
int head[N], pre[N][N], Sum[N], d[N];
bool vis[N];

void AddEdge(int u, int v, int dist) {
    E[tot].to = v;
    E[tot].dist = dist;
    E[tot].next = head[u];
    head[u] = tot++;
}

void init() {
    memset(head, -1, sizeof(head));
    tot = 0;

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            if (i != j)
                AddEdge(i, j, l);
        }

    int u, v, d;
    for (int i = 0; i < m; i++){
        scanf("%d%d%d", &u, &v, &d);
        node[i].x = u;
        node[i].y = v;
        node[i].id = tot;
        AddEdge(u, v, d);
        AddEdge(v, u, d);
    }
}

void SPFA1(int s) {
    for (int i = 1; i <= n; i++) {
        d[i] = INF;
        vis[i] = false;
    }

    queue<int> q;
    q.push(s);
    d[s] = 0;
    pre[s][s] = 0;

    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = false;

        for (int i = head[u]; i != -1; i = E[i].next) {
            int v = E[i].to;
            if (d[v] > d[u] + E[i].dist) {
                d[v] = d[u] + E[i].dist;
                pre[s][v] = u;
                if (!vis[v]) {
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }

    for (int i = 1; i <= n; i++)
        Sum[s] += d[i];
}

int SPFA2(int s, int id) {
    for (int i = 1; i <= n; i++) {
        d[i] = INF;
        vis[i] =false;
    }
    d[s] = 0;
    queue<int> q;
    q.push(s);

    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for (int i = head[u]; i != -1; i = E[i].next) {
            if (i == id || i == id + 1)
                continue;
            int v = E[i].to;
            if (d[v] > d[u] + E[i].dist) {
                d[v] = d[u] + E[i].dist;
                if (!vis[v]) {
                    q.push(v);
                    vis[v] = true;
                }
            }
        }
    }

    int S = 0;
    for (int i = 1; i <= n; i++)
        S += d[i];

    return S;
}

void solve() {
    int S = 0;
    for (int i = 1; i <= n; i++) {
        Sum[i] = 0;
        SPFA1(i);
        S += Sum[i];
    }

    int Max = -INF;
    for (int i = 0; i < m; i++) {
        int t = 0;
        int x = node[i].x, y = node[i].y, id = node[i].id;
        for (int j = 1; j <= n; j++) {
            if (pre[j][x] != y && pre[j][y] != x) {
                t += Sum[j];
                continue;
            }
            t += SPFA2(j, id);
        }
        Max = max(Max, t);
    }
    printf("%d %d\n", S, Max);
}

int main() {
    while (scanf("%d%d%d", &n, &m, &l) != EOF) {
        init();
        solve();
    }
    return 0;
}

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