1. 技术文章
  2. 【代码超详解】POJ 1511 / ZOJ 2008 Invitation Cards（vector 存图，Dijkstra 算法，思维）

【代码超详解】POJ 1511 / ZOJ 2008 Invitation Cards（vector 存图，Dijkstra 算法，思维）

由 谁都会走 提交于 2020-03-10 05:32:48

传送门：
http://poj.org/problem?id=1511
https://zoj.pintia.cn/problem-sets/91827364500/problems/91827365507

一、题目描述

Invitation Cards

Time Limit: 5000 ms（ZOJ），8000 ms（POJ）
Memory Limit: 65536 KB（ZOJ），262144 KB（POJ）

In the age of television, not many people attend theater performances. Antique Comedians of Malidinesia are aware of this fact. They want to propagate theater and, most of all, Antique Comedies. They have printed invitation cards with all the necessary information and with the programme. A lot of students were hired to distribute these invitations among the people. Each student volunteer has assigned exactly one bus stop and he or she stays there the whole day and gives invitation to people travelling by bus. A special course was taken where students learned how to influence people and what is the difference between influencing and robbery.

The transport system is very special: all lines are unidirectional and connect exactly two stops. Buses leave the originating stop with passangers each half an hour. After reaching the destination stop they return empty to the originating stop, where they wait until the next full half an hour, e.g. X:00 or X:30, where ‘X’ denotes the hour. The fee for transport between two stops is given by special tables and is payable on the spot. The lines are planned in such a way, that each round trip (i.e. a journey starting and finishing at the same stop) passes through a Central Checkpoint Stop (CCS) where each passenger has to pass a thorough check including body scan.

All the ACM student members leave the CCS each morning. Each volunteer is to move to one predetermined stop to invite passengers. There are as many volunteers as stops. At the end of the day, all students travel back to CCS. You are to write a computer program that helps ACM to minimize the amount of money to pay every day for the transport of their employees.

Input

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case begins with a line containing exactly two integers P and Q, 1 <= P,Q <= 1000000. P is the number of stops including CCS and Q the number of bus lines. Then there are Q lines, each describing one bus line. Each of the lines contains exactly three numbers - the originating stop, the destination stop and the price. The CCS is designated by number 1. Prices are positive integers the sum of which is smaller than 1000000000. You can also assume it is always possible to get from any stop to any other stop.

Output

For each case, print one line containing the minimum amount of money to be paid each day by ACM for the travel costs of its volunteers.

Sample Input

2
2 2
1 2 13
2 1 33
4 6
1 2 10
2 1 60
1 3 20
3 4 10
2 4 5
4 1 50

Sample Output

46
210

二、算法分析说明与代码编写指导

求：顶点 1 到其它各顶点的最小花费与各顶点返回顶点 1 的最小花费。
建议先阅读方法一致的 POJ 3268 题解。

对本题而言：存图如果用同时存正反两张，POJ 上即使用快读也超时，ZOJ 则直接爆内存。
POJ 的 vector 抹了猪油，同样的代码放在 ZOJ 上只用 321 ms，而 POJ 上超过 7100 ms（用快读降低到约 6100 ms）。而 POJ 关于此问题的讨论区有大量反映用 vector 存图超时的情况。另外 POJ 上用 C++ 提交，如果用了快读反而还会超时，非常玄学。
ZOJ 上提交本题，将所有的 push 换成 emplace，时间上几乎没有区别（321 ms vs 327 ms，理论上 emplace 更快）
这题巨坑，明明说了所有价格加起来不超过 1e9，结果还是超了。如果不超的话 unsigned 的范围也够了，但实际上提交上去是 WA 的。

三、AC 代码

POJ：

#include<cstdio>
#include<vector>
#include<algorithm>
#include<bitset>
#include<queue>
#pragma warning(disable:4996)
using namespace std;
const unsigned long long Pmax = 1 << 30; const unsigned Qmax = 1e6 + 1;
struct edge { unsigned v; unsigned long long p; }; struct Edge { unsigned u, v; unsigned long long p; };
unsigned N, P, Q, x, y; vector<edge> G[Qmax]; edge e; Edge E[Qmax]; bitset<Qmax> v;
unsigned long long* p, p1[Qmax], p2[Qmax], a;
struct pred {
	bool operator()(const unsigned& l, const unsigned& r) { return p[l] > p[r]; }
};
priority_queue<unsigned, vector<unsigned>, pred> q;
inline void dijkstra() {
	fill(p + 1, p + P + 1, Pmax); p[1] = 0; v.reset(); q.push(1);
	while (!q.empty()) {
		x = q.top(); q.pop(); if (v[x])continue;
		v[x] = true;
		for (size_t i = 0; i < G[x].size(); ++i) {
			y = G[x][i].v; p[y] = min(p[y], p[x] + G[x][i].p); q.push(y);
		}
	}
}
int main() {
	scanf("%u", &N); ++N;
	while (--N) {
		scanf("%u%u", &P, &Q); a = 0;
		for (unsigned i = 1; i <= P; ++i)G[i].clear();
		for (unsigned i = 1; i <= Q; ++i) {
			scanf("%u%u%llu", &E[i].u, &E[i].v, &E[i].p); e.v = E[i].v; e.p = E[i].p; G[E[i].u].push_back(e);
		}
		p = p1; dijkstra();
		for (unsigned i = 1; i <= P; ++i)G[i].clear();
		for (unsigned i = 1; i <= Q; ++i) { e.v = E[i].u; e.p = E[i].p; G[E[i].v].push_back(e); }
		p = p2; dijkstra();
		for (unsigned i = 1; i <= P; ++i)a += p1[i] + p2[i];
		printf("%llu\n", a);
	}
	return 0;
}

POJ：（快读版本）

#include<cstdio>
#include<vector>
#include<algorithm>
#include<bitset>
#include<queue>
#include<cctype>
#pragma warning(disable:4996)
using namespace std;
const unsigned long long Pmax = 1 << 30; const unsigned Qmax = 1e6 + 1;
struct edge { unsigned v; unsigned long long p; }; struct Edge { unsigned u, v; unsigned long long p; };
unsigned N, P, Q, x, y; vector<edge> G[Qmax]; edge e; Edge E[Qmax]; bitset<Qmax> v;
unsigned long long* p, p1[Qmax], p2[Qmax], a;
struct pred {
	bool operator()(const unsigned& l, const unsigned& r) { return p[l] > p[r]; }
};
priority_queue<unsigned, vector<unsigned>, pred> q;
inline void dijkstra() {
	fill(p + 1, p + P + 1, Pmax); p[1] = 0; v.reset(); q.push(1);
	while (!q.empty()) {
		x = q.top(); q.pop(); if (v[x])continue;
		v[x] = true;
		for (size_t i = 0; i < G[x].size(); ++i) {
			y = G[x][i].v; p[y] = min(p[y], p[x] + G[x][i].p); q.push(y);
		}
	}
}
template<class _Ty> inline void readu(_Ty& x) {
	static _Ty c; x = 0;
	for (;;) {
		c = getchar();
		if (!isdigit(c))return;
		x = x * 10 + c - '0';
	}
}
int main() {
	readu(N); ++N;
	while (--N) {
		readu(P); readu(Q); a = 0;
		for (unsigned i = 1; i <= P; ++i)G[i].clear();
		for (unsigned i = 1; i <= Q; ++i) {
			readu(E[i].u); readu(E[i].v); readu(E[i].p); e.v = E[i].v; e.p = E[i].p; G[E[i].u].push_back(e);
		}
		p = p1; dijkstra();
		for (unsigned i = 1; i <= P; ++i)G[i].clear();
		for (unsigned i = 1; i <= Q; ++i) { e.v = E[i].u; e.p = E[i].p; G[E[i].v].push_back(e); }
		p = p2; dijkstra();
		for (unsigned i = 1; i <= P; ++i)a += p1[i] + p2[i];
		printf("%llu\n", a);
	}
	return 0;
}

ZOJ：

#include<cstdio>
#include<vector>
#include<algorithm>
#include<bitset>
#include<queue>
#pragma warning(disable:4996)
using namespace std;
const unsigned long long Pmax = 1 << 30; const unsigned Qmax = 1e6 + 1;
struct edge { unsigned v; unsigned long long p; }; struct Edge { unsigned u, v; unsigned long long p; };
unsigned N, P, Q, x, y; vector<edge> G[Qmax]; edge e; Edge E[Qmax]; bitset<Qmax> v;
unsigned long long* p, p1[Qmax], p2[Qmax], a;
struct pred {
	bool operator()(const unsigned& l, const unsigned& r) { return p[l] > p[r]; }
};
priority_queue<unsigned, vector<unsigned>, pred> q;
inline void dijkstra() {
	fill(p + 1, p + P + 1, Pmax); p[1] = 0; v.reset(); q.emplace(1);
	while (!q.empty()) {
		x = q.top(); q.pop(); if (v[x])continue;
		v[x] = true;
		for (size_t i = 0; i < G[x].size(); ++i) {
			y = G[x][i].v; p[y] = min(p[y], p[x] + G[x][i].p); q.emplace(y);
		}
	}
}
int main() {
	scanf("%u", &N); ++N;
	while (--N) {
		scanf("%u%u", &P, &Q); a = 0;
		for (unsigned i = 1; i <= P; ++i)G[i].clear();
		for (unsigned i = 1; i <= Q; ++i) {
			scanf("%u%u%llu", &E[i].u, &E[i].v, &E[i].p); e.v = E[i].v; e.p = E[i].p; G[E[i].u].emplace_back(e);
		}
		p = p1; dijkstra();
		for (unsigned i = 1; i <= P; ++i)G[i].clear();
		for (unsigned i = 1; i <= Q; ++i) { e.v = E[i].u; e.p = E[i].p; G[E[i].v].emplace_back(e); }
		p = p2; dijkstra();
		for (unsigned i = 1; i <= P; ++i)a += p1[i] + p2[i];
		printf("%llu\n", a);
	}
	return 0;
}
标签
each
易学教程内所有资源均来自网络或用户发布的内容，如有违反法律规定的内容欢迎反馈！
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!