【BZOJ】1681: [Usaco2005 Mar]Checking an Alibi 不在场的证明(spfa)



#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }
const int N=505, Q=N*10, oo=~0u>>2;
int q[Q], front, tail, d[N], vis[N], ihead[N], cnt, n, x, y, m, ans;
struct ED { int to, w, next; }e[Q];
void add(int u, int v, int w) {
    e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].to=v; e[cnt].w=w;
    e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].to=u; e[cnt].w=w;
void spfa() {
    for1(i, 2, n) d[i]=oo;
    while(front!=tail) {
        int v, u=q[front++]; if(front==Q) front=0; vis[u]=0;
        for(int i=ihead[u]; i; i=e[i].next) if(d[v=e[i].to]>d[u]+e[i].w) {
            if(!vis[v]) {
                q[tail++]=v; if(tail==Q) tail=0;
int main() {
    read(n); read(m); read(x); read(y);
    for1(i, 1, m) {
        int u=getint(), v=getint(), w=getint();
        add(u, v, w);
    CC(vis, 0);
    for1(i, 1, x) {
    	int t=getint();
    	if(d[t]<=y) vis[i]=1, ++ans;
    printf("%d\n", ans);
    for1(i, 1, x) if(vis[i]) printf("%d\n", i);
    return 0;






A crime has been comitted: a load of grain has been taken from the barn by one of FJ's cows. FJ is trying to determine which of his C (1 <= C <= 100) cows is the culprit. Fortunately, a passing satellite took an image of his farm M (1 <= M <= 70000) seconds before the crime took place, giving the location of all of the cows. He wants to know which cows had time to get to the barn to steal the grain. Farmer John's farm comprises F (1 <= F <= 500) fields numbered 1..F and connected by P (1 <= P <= 1,000) bidirectional paths whose traversal time is in the range 1..70000 seconds (cows walk very slowly). Field 1 contains the barn. It takes no time to travel within a field (switch paths). Given the layout of Farmer John's farm and the location of each cow when the satellite flew over, determine set of cows who could be guilty. NOTE: Do not declare a variable named exactly 'time'. This will reference the system call and never give you the results you really want.

    约翰农场有F(1≤F≤500)草地,标号1到F,还有P(1≤P≤1000)条双向路连接着它们.通过这些路需要的时间在1到70000秒的范围内.田地1上建有那个被盗的谷仓. 给出农场地图,以及卫星照片里每只牛所在的位置.请判断哪些牛有可能犯罪.


* Line 1: Four space-separated integers: F, P, C, and M * Lines 2..P+1: Three space-separated integers describing a path: F1, F2, and T. The path connects F1 and F2 and requires T seconds to traverse. * Lines P+2..P+C+1: One integer per line, the location of a cow. The first line gives the field number of cow 1, the second of cow 2, etc.





* Line 1: A single integer N, the number of cows that could be guilty of the crime.

* Lines 2..N+1: A single cow number on each line that is one of the cows that could be guilty of the crime. The list must be in ascending order.


Sample Input

7 6 5 8
1 4 2
1 2 1
2 3 6
3 5 5
5 4 6
1 7 9

Sample Output




posted @ 2014-09-14 00:14  iwtwiioi  阅读(269)  评论(0编辑  收藏  举报