Codeforces Round #103 (Div. 2) D题 SPFA

题意就是求离源点距离为L的点的个数，点可以在节点上，也可以在路上，但是必须都是到源点的最短距离为L

用SPFA求一遍距离，然后扫描一遍点，再扫描一遍边

扫描边得时候注意了，有的边上有1个位置，有的边上有2个位置，并且要去重

/*ID: sdj22251PROG: subsetLANG: C++*/#include <iostream>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cctype>#include <string>#include <cstring>#include <cmath>#include <ctime>#define LOCA#define MAXN 100005#define INF 100000005#define eps 1e-7using namespace std;struct node{    int v, w;    node *next;} edge[MAXN], temp[3 * MAXN]; int d[MAXN], n, m, pos = 0;int q[MAXN * 4];bool visited[MAXN]; int uu[MAXN], vv[MAXN], ww[MAXN];void spfa(int src, int *ecost) //src是起点， ecost是边权{    int h, t, u, i;    node *ptr;    h = 0, t = 1;    memset(visited, 0, sizeof(visited));    q[0] = src;    ecost[src] = 0;    visited[src] = true;    while(h < t)    {        u = q[h++];        visited[u] = false;        ptr = edge[u].next;        while(ptr)        {            if(ecost[ptr -> v] > ecost[u] + ptr -> w)            {                ecost[ptr -> v] = ecost[u] + ptr -> w;                if(!visited[ptr -> v])                {                    q[t++] = ptr -> v;                    visited[ptr -> v] = true;                }            }            ptr = ptr -> next;        }    }}void insert(const int &x, const int &y, const int &w){    node *ptr = &temp[pos++];     ptr -> v = y;    ptr -> w = w;    ptr -> next = edge[x].next;    edge[x].next = ptr;}void init(){    for(int i = 0; i <= n; i++)    {        edge[i].next = NULL;        d[i] = INF;    }}int main(){    int s, w, len;    scanf("%d%d%d", &n, &m, &s);    init();    for(int i = 0; i < m; i++)    {        scanf("%d%d%d", &uu[i], &vv[i], &ww[i]);        insert(uu[i], vv[i], ww[i]);        insert(vv[i], uu[i], ww[i]);    }    scanf("%d", &len);    spfa(s, d);    int ans = 0;    for(int i = 1; i <= n; i++)    {        if(d[i] == len)        ans++;    }    for(int i = 0; i < m; i++)    {        int mi = min(d[uu[i]], d[vv[i]]);        int mx = max(d[uu[i]], d[vv[i]]);        if(mi < len && mx < len) //如果两个点到源点的最短距离都小于L，就有可能出现边上有两个位置符合题意        {            if(mi + mx + ww[i] < 2 * len)       //由于题目要求是到源点的最短距离为L，那么两点分到源点的最短距离之和加上边权如果小于2*L，显然任何位置的最短距离都是小于L的            continue;            if(mi + mx + ww[i] == 2 * len) //去重，当某个位置通过两个结点到到达的源点都是L的时候            ans++;            else if(mi + mx + ww[i] > 2 * len)            ans += 2;        }        else if(mi < len)        {            if(mi + ww[i] > len)            ans++;        }    }    printf("%d\n", ans);    return 0;}