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

最新推荐文章于 2020-03-07 16:44:12 发布

转载最新推荐文章于 2020-03-07 16:44:12 发布 · 176 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/iwtwiioi/p/3970552.html

文章标签：

#php

本文深入探讨了SPFA算法的实现细节，包括其在解决最短路径问题上的应用，特别是在处理具有多个源点和复杂边权的图论问题中。通过对一个具体的编程竞赛题目进行分析，展示了如何使用SPFA算法高效地找到所有可能的罪犯奶牛，为读者提供了丰富的代码实例和算法解释。

http://www.lydsy.com/JudgeOnline/problem.php?id=1681

太裸了。。

#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() {
    q[tail++]=1;
    for1(i, 2, n) d[i]=oo;
    vis[1]=1;
    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) {
            d[v]=d[u]+e[i].w;
            if(!vis[v]) {
                vis[v]=1;
                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);
    }
    spfa();
    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;
}

Description

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.

谷仓里发现谷物被盗！约翰正试图从C(1≤C≤100)只奶牛里找出那个偷谷物的罪犯．幸运的是，一个恰好路过的卫星拍下谷物被盗前M(1≤M≤70000)秒的农场的图片．这样约翰就能通过牛们的位置来判断谁有足够的时间来盗窃谷物．

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．

Input

* 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.

第1行输入四个整数F，只C，和M;

接下来P行每行三个整数描述一条路，起点终点和通过时间．

接下来C行每行一个整数，表示一头牛所在的地点．