2015 四川省赛 I Travel（bfs）

题意:

给定N<=105的无向完全图,给出其中M<=5×105条边,边权为a,其余边权为b,求1−>n最短路

分析:

分类讨论即可:

a<b时如果1−>n在给出的图上那么答案显然是a,反之那么1−>n边在当前图的补图上,答案就是min(当前图的dis[n]∗a,b)

a>=b时如果1−>n在给出的图的补图上那么答案显然是b,反之那么1−>n边在当前图上,答案就是min(补图的dis[n]∗b,a)

关于补图bfs的时候由于图巨大,不能存储,我们可以con[v]=u判断当前点所连的边是不是在原图上,在就跳过这个边

同时为了保证复杂度我们用set来维护剩余没有得到最短路的顶点集合,这样就保证遍历的复杂度是O(max(n,m))了

问题得以解决

坑: set遍历的时候,有删除操作红黑树会调整,所以需要重新查找下一个值来继续遍历

代码:

//
//  Created by TaoSama on 2015-10-01
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int M = 1e6 + 10;

int n, m, a, b, T;
bool des
;

int head
, pnt[M], nxt[M], cnt;
void add_edge(int u, int v) {
pnt[cnt] = v;
nxt[cnt] = head[u];
head[u] = cnt++;
}

int dp
, con
;
int bfs() {
queue<int> q;
memset(dp, 0x3f, sizeof dp);
dp[1] = 0; q.push(1);
while(q.size()) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = nxt[i]) {
int v = pnt[i];
if(dp[v] == INF) {
dp[v] = dp[u] + 1;
if(v == n) return dp[v];
q.push(v);
}
}
}
return dp
;
}

int cbfs() {
queue<int> q;
memset(dp, 0x3f, sizeof dp);
memset(con, 0, sizeof con);
set<int> s;
for(int i = 2; i <= n; ++i) s.insert(i);
dp[1] = 0; q.push(1);
while(q.size()) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = nxt[i]) con[pnt[i]] = u;
int v = -1;
while(true) {
set<int>::iterator iter = s.upper_bound(v);
if(iter == s.end()) break;
v = *iter;
if(con[v] != u) {
s.erase(iter);
dp[v] = dp[u] + 1;
if(v == n) return dp[v];
q.push(v);
}
}
}
return dp
;
}

inline int read() {
char c = getchar();
while(!isdigit(c)) c = getchar();
int x = 0;
while(isdigit(c)) {
x = x * 10 + c - '0';
c = getchar();
}
return x;
}

int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
//  freopen("out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

while(scanf("%d%d%d%d", &n, &m, &a, &b) == 4) {
bool ori = false;
cnt = 0; memset(head, -1, sizeof head);
memset(des, false, sizeof des);
for(int i = 1; i <= m; ++i) {
int u = read(), v = read();
if(u > v) swap(u, v);
add_edge(u, v);
add_edge(v, u);
if(v == n) des[u] = true;
if(u == 1 && v == n) ori = true;
}
long long ans;
if(a < b) {
if(ori) ans = a;
else {
ans = min(0LL + b, 1LL * a * bfs());
}
} else {
if(!ori) ans = b;
else {
ans = min(0LL + a, 1LL * b * cbfs());
}
}
printf("%lld\n", ans);
}
return 0;
}