第五十七题 UVa 1416 or LA 4080 Warfare And Logistics

传送门

有、艹蛋
用来复习Dijkstra算法啦

#include <cstdio>  
#include <cstring>  
#include <algorithm>  
#include <vector>  
#include <queue>  
using namespace std;  
const int INF = 0x7fffffff;  
const int maxn = 100 + 10;  
struct Edge{  
    int from, to, dist;  
};  
struct HeapNode{  
    int d, u;  
    bool operator < (const HeapNode& rhs) const {  
        return d > rhs.d;  
    }  
};  
struct Dijkstra{  
    int n, m;  
    vector<Edge> edges;  
    vector<int> G[maxn];  
    bool done[maxn];  
    int p[maxn], d[maxn];  
    void init(int n){  
        this -> n = n;  
        for (int i = 0; i < n; i++)  
            G[i].clear();  
        edges.clear();  
    }  
    void AddEdge(int from, int to, int dist){  
        edges.push_back((Edge){from, to, dist});  
        m = edges.size();  
        G[from].push_back(m - 1);  
    }  
    void dijkstra(int s){  
        priority_queue<HeapNode> que;  
        for (int i = 0; i < n; i++)  
            d[i] = INF;  
        d[s] = 0;  
        memset(done, 0, sizeof(done));  
        que.push((HeapNode){0, s});  
        while(!que.empty()){  
            HeapNode x = que.top(); que.pop();  
            int u = x.u;  
            if (done[u]) continue;  
            done[u] = 1;  
            for (int i = 0; i < G[u].size(); i++){  
                Edge& e = edges[G[u][i]];  
                if (e.dist >= 0 && d[e.to] > d[u] + e.dist){  
                    d[e.to] = d[u] + e.dist;  
                    p[e.to] = G[u][i];  
                    que.push((HeapNode){d[e.to], e.to});  
                }  
            }  
        }  
    }  
};  
Dijkstra solver;  
int n, m, L;  
vector<int> gr[maxn][maxn];  
int used[maxn][maxn][maxn];  
int idx[maxn][maxn];  
int sum_single[maxn];  
int compute_c(){  
    int ans = 0;  
    memset(used, 0, sizeof(used));  
    for (int src = 0; src < n; src++){  
        solver.dijkstra(src);  
        sum_single[src] = 0;  
        for (int i = 0; i < n; i++){  
            if (i != src){  
                int fa = solver.edges[solver.p[i]].from;  
                used[src][fa][i] = used[src][i][fa] = 1;  
            }  
            sum_single[src] += (solver.d[i] == INF ? L : solver.d[i]);  
        }  
        ans += sum_single[src];  
    }  
    return ans;  
}  
int compute_newc(int a, int b){  
    int ans = 0;  
    for (int src = 0; src < n; src++)  
        if (!used[src][a][b]) ans += sum_single[src];  
        else{  
            solver.dijkstra(src);  
            for (int i = 0; i < n; i++)  
                ans += (solver.d[i] == INF ? L : solver.d[i]);  
        }  
    return ans;  
}  
int main(){  
//  freopen("in.txt", "r", stdin);  
    while(~scanf("%d%d%d", &n, &m, &L)){  
        solver.init(n);  
        for (int i = 0; i < n; i++)  
            for (int j = 0; j < n; j++)  
                gr[i][j].clear();  
        for (int i = 0; i < m; i++){  
            int x, y, z;  
            scanf("%d%d%d", &x, &y, &z);  
            x -= 1; y -= 1;  
            gr[x][y].push_back(z);  
            gr[y][x].push_back(z);  
        }  
        for (int i = 0; i < n; i++)  
            for (int j = i + 1; j < n; j++)  
                if (!gr[i][j].empty()){  
                    sort(gr[i][j].begin(), gr[i][j].end());  
                    solver.AddEdge(i, j, gr[i][j][0]);  
                    idx[i][j] = solver.m - 1;  
                    solver.AddEdge(j, i, gr[i][j][0]);  
                    idx[j][i] = solver.m - 1;  
                }  
        int c = compute_c(), c2 = -1;  
        for (int i = 0; i < n; i++)  
            for (int j = i + 1; j < n; j++)  
                if (!gr[i][j].empty()){  
                    int& e1 = solver.edges[idx[i][j]].dist;  
                    int& e2 = solver.edges[idx[j][i]].dist;  
                    if (gr[i][j].size() == 1) e1 = e2 = -1;  
                    else e1 = e2 = gr[i][j][1];  
                    c2 = max(c2, compute_newc(i, j));  
                    e1 = e2 = gr[i][j][0];  
                }  
        printf("%d %d\n", c, c2);  
    }  
    return 0;  
}