UVA1416 Warfare And Logistics(LA4080)

首先用Dijkstra求出每对点距离；
以及每条边是否属于以某个点为单源的树上


然后枚举删除每条边:
之后对于每个单源，如果这条边在上面则重新对这个点dijkstra,
然后修改相应的边；

否则每个点到这个单源的最短距离肯定不变(在原C基础上修改)

wrong了整整一天，原来是求与源点相邻的点时忘了会有重边的情况，坑死了……

#include <stdio.h>#include <string.h>#include <vector>#include <queue>using namespace std;typedef long long LL;const int maxn = 100+10;const int maxm = 1000+10;class Edge{    public:    int pos, no;//no表示它来自的边的编号    LL dis;//邻接表的边长或优先队列中到源点距离    Edge(int pos, LL dis, int no = 0)    {        this->pos = pos;        this->dis = dis;        this->no  = no;    }    bool operator < (const Edge &ans)const    {        return dis > ans.dis;    }};vector<Edge> adj[maxn];int dis[maxn][maxn], dis1[maxn];bool belong[maxm][maxn];//某边是否在某棵树上bool vis[maxn], del[maxm];//del[i]这条边是否被删除了int N, M, L;LL C, CC, CC1;

void Init(){    for(int i = 1; i <= N; i++)        adj[i].clear();    memset(belong, false, sizeof(belong));    memset(del, false, sizeof(del));    int a, b, c;    for(int i = 1; i <= M; i++)    {        scanf("%d%d%d", &a, &b, &c);        adj[a].push_back(Edge(b, c, i));        adj[b].push_back(Edge(a, c, i));    }}void Dijkstra1(int dd[], int root){    for(int i = 1; i <= N; i++)    {        vis[i] = false;        dd[i] = -1;    }    dd[root] = 0;    vis[root] = true;    priority_queue<Edge> myQue;    for(vector<Edge>::iterator it = adj[root].begin(); it != adj[root].end(); it++)    {        int u = it->pos;        if(dd[u] == -1 || it->dis < dd[u])//可能有重边，就是这个地方让我wrong了一整天啊        {            dd[u] = it->dis;            myQue.push(Edge(u, dd[u], it->no));        }    }    while(!myQue.empty())    {        Edge ans = myQue.top();        myQue.pop();        int u = ans.pos;        if(vis[u])            continue;        vis[u] = true;        belong[ans.no][root] = true;//ans.no属于以root为根的树        for(vector<Edge>::iterator it = adj[u].begin(); it != adj[u].end(); it++)        {            int v = it->pos;            if(dd[v] == -1 || dd[u] + it->dis < dd[v])            {                dd[v] = dd[u] + it->dis;                myQue.push(Edge(v, dd[v], it->no));            }        }    }    for(int i = 1; i <= N; i++)        if(dd[i] == -1)            C += L;        else            C += dd[i];}void Dijkstra2(int dd[], int root){    for(int i = 1; i <= N; i++)    {        dd[i] = -1;        vis[i] = false;    }    priority_queue<Edge> myQue;    dd[root] = 0;    myQue.push(Edge(root, 0));//no在进入队列后就没用了,但是进入前要判断    while(!myQue.empty())    {        Edge ans = myQue.top();        myQue.pop();        int u = ans.pos;        if(vis[u])            continue;        vis[u] = true;        for(vector<Edge>::iterator it = adj[u].begin(); it != adj[u].end(); it++)        {            if(del[it->no])//这条边被删除了                continue;            int v = it->pos;            if(dd[v] == -1 || dd[u] + it->dis < dd[v])            {                dd[v] = dd[u] + it->dis;                myQue.push(Edge(v, dd[v]));            }        }    }    for(int i = 1; i <= N; i++)//替换相应的边    {        if(dis[root][i] == -1)            CC1 -= L;        else            CC1 -= dis[root][i];        if(dd[i] == -1)            CC1 += L;        else            CC1 += dd[i];    }}void Solve(){    C = 0;    for(int i = 1; i <= N; i++)        Dijkstra1(dis[i], i);    CC = C;    for(int i = 1; i <= M; i++)//枚举删除了这条边    {        CC1 = C;//只在原来不删除边的基础上修改变化的边        del[i] = true;//删除这条边        for(int j = 1; j <= N; j++)            if(belong[i][j])//i边在以j为根的树上                Dijkstra2(dis1, j);        del[i] = false;        if(CC == -1 || CC < CC1)            CC = CC1;    }    printf("%lld %lld\n", C, CC);}int main(){    while(scanf("%d%d%d", &N, &M, &L) != EOF)    {        Init();        Solve();    }    return 0;}