洛谷 P3003 [USACO10DEC]蘋果交貨Apple Delivery

　　　　　　　洛谷 P3003 [USACO10DEC]蘋果交貨Apple Delivery

題目描述

Bessie has two crisp red apples to deliver to two of her friends in the herd. Of course, she travels the C (1 <= C <= 200,000)

cowpaths which are arranged as the usual graph which connects P (1 <= P <= 100,000) pastures conveniently numbered from 1..P: no cowpath leads from a pasture to itself, cowpaths are bidirectional, each cowpath has an associated distance, and, best of all, it is always possible to get from any pasture to any other pasture. Each cowpath connects two differing pastures P1_i (1 <= P1_i <= P) and P2_i (1 <= P2_i <= P) with a distance between them of D_i. The sum of all the distances D_i does not exceed 2,000,000,000.

What is the minimum total distance Bessie must travel to deliver both apples by starting at pasture PB (1 <= PB <= P) and visiting pastures PA1 (1 <= PA1 <= P) and PA2 (1 <= PA2 <= P) in any order. All three of these pastures are distinct, of course.

Consider this map of bracketed pasture numbers and cowpaths with distances:

3 2 2 [1]-----[2]------[3]-----[4] \ / \ / 7\ /4 \3 /2 \ / \ / [5]-----[6]------[7] 1 2

If Bessie starts at pasture [5] and delivers apples to pastures [1] and [4], her best path is:

5 -> 6-> 7 -> 4* -> 3 -> 2 -> 1*

with a total distance of 12.

貝西有兩個又香又脆的紅蘋果要送給她的兩個朋友。固然她能夠走的C（1<=C<=200000）條「牛路」都被包含在一種經常使用的圖中，包含了P（1<=P<=100000）個牧場，分別被標爲1..P。沒有「牛路」會從一個牧場又走回它本身。「牛路」是雙向的，每條牛路都會被標上一個距離。最重要的是，每一個牧場均可以通向另外一個牧場。每條牛路都鏈接着兩個不一樣的牧場P1_i和P2_i（1<=P1_i,p2_i<=P)，距離爲D_i。全部「牛路」的距離之和不大於2000000000。

如今，貝西要從牧場PB開始給PA_1和PA_2牧場各送一個蘋果（PA_1和PA_2順序能夠調換），那麼最短的距離是多少呢？固然，PB、PA_1和PA_2各不相同。

輸入輸出格式

輸入格式：

* Line 1: Line 1 contains five space-separated integers: C, P, PB, PA1, and PA2

* Lines 2..C+1: Line i+1 describes cowpath i by naming two pastures it connects and the distance between them: P1_i, P2_i, D_i

輸出格式：

* Line 1: The shortest distance Bessie must travel to deliver both apples

輸入輸出樣例

輸入樣例#1：  複製

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

輸出樣例#1：  複製

12 

思路：SPFA + SLF 優化（之前歷來沒使過）        難度：提升+/省選- （自認爲難度應該再低點）
接下來說一下我作這道題的（被坑）歷程
首先我看了看這道題 不就是跑兩邊SPFA嗎，而後。。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define M 200005
#define MAXN 0x7fffffff
using namespace std;
queue<int> q;
int m, n, s, t1, t2;
int tot, minn;
int dis[M], vis[M];
int to[M*2], net[M*2], head[M*2], cap[M*2];

void add(int u, int v, int w) {
    to[++tot] = v; net[tot] = head[u]; head[u] = tot; cap[tot] = w;
    to[++tot] = u; net[tot] = head[v]; head[v] = tot; cap[tot] = w;
}

void spfa(int x) {
    for(int i = 1; i <= n; i++) vis[i] = 0, dis[i] = MAXN;
    dis[x] = 0; vis[x] = 1; q.push(x);
    while(!q.empty()) {
        int y = q.front(); q.pop(); vis[y] = 0;
        for(int i = head[y]; i; i = net[i]) {
            int t = to[i];
            if(dis[t] > dis[y]+cap[i]) {
                dis[t] = dis[y]+cap[i];
                if(!vis[t]) vis[t] = 1, q.push(t);
            }
        }
    }
}

int main() {
    scanf("%d%d%d%d%d", &m, &n, &s, &t1, &t2);
    for(int i = 1; i <= m; i++) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }
    spfa(t1);
    minn = dis[s] + dis[t2];
    spfa(t2);
    minn = min(minn, dis[s]+dis[t1]);
    printf("%d", minn);
    return 0;
}

裸的SPFA，RE了兩個點

竟然RE了！我應該開的夠大啊，而後數組多開了一個0，通通開long long，and then。。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define LL long long
#define M 2000005
#define MAXN 0x7fffffff
using namespace std;
queue<LL> q;
LL m, n, s, t1, t2;
LL tot, minn;
LL dis[M], vis[M];
LL to[M*2], net[M*2], head[M*2], cap[M*2];

void add(LL u, LL v, LL w) {
    to[++tot] = v; net[tot] = head[u]; head[u] = tot; cap[tot] = w;
    to[++tot] = u; net[tot] = head[v]; head[v] = tot; cap[tot] = w;
}

void spfa(LL x) {
    for(LL i = 1; i <= n; i++) vis[i] = 0, dis[i] = MAXN;
    dis[x] = 0; vis[x] = 1; q.push(x);
    while(!q.empty()) {
        int y = q.front(); q.pop(); vis[y] = 0;
        for(LL i = head[y]; i; i = net[i]) {
            LL t = to[i];
            if(dis[t] > dis[y]+cap[i]) {
                dis[t] = dis[y]+cap[i];
                if(!vis[t]) vis[t] = 1, q.push(t);
            }
        }
    }
}

int main() {
    scanf("%lld%lld%lld%lld%lld", &m, &n, &s, &t1, &t2);
    for(LL i = 1; i <= m; i++) {
        LL a, b, c;
        scanf("%lld%lld%lld", &a, &b, &c);
        add(a, b, c);
    }
    spfa(t1);
    minn = dis[s] + dis[t2];
    spfa(t2);
    minn = min(minn, dis[s]+dis[t1]);
    printf("%lld", minn);
    return 0;
} 

結果並無什麼卵用，繼續RE兩個點。。

而後我明智（實在沒辦法）的問了學姐，結果她告訴我：「這個題我記得要用SLF優化」 QAQ

可是我不會啊

而後學姐講了加了SLF以後的變化，可是不知道爲啥，我樣例也過不了了 （大哭）

經過比較發現，不開long long的時候結果仍是對滴。。

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<deque>
#define M 200005
#define MAXN 0x7fffffff
using namespace std;
deque<int> q;
int m, n, s, t1, t2;
int tot, minn;
int dis[M], vis[M];
int to[M*2], net[M*2], head[M*2], cap[M*2];

void add(int u, int v, int w) {
    to[++tot] = v; net[tot] = head[u]; head[u] = tot; cap[tot] = w;
    to[++tot] = u; net[tot] = head[v]; head[v] = tot; cap[tot] = w;
}

void spfa(int x) {
    for(int i = 1; i <= n; i++) vis[i] = 0, dis[i] = MAXN;
    dis[x] = 0; vis[x] = 1; q.push_back(x);
    while(!q.empty()) {
        int y = q.front(); q.pop_front(); vis[y] = 0;
        for(int i = head[y]; i; i = net[i]) {
            int t = to[i];
            if(dis[t] > dis[y]+cap[i]) {
                dis[t] = dis[y]+cap[i];
                if(!vis[t]) {
                    vis[t] = 1;
                    if(q.empty() || dis[t]<dis[q.front()]) q.push_front(t);
                    else q.push_back(t);
                }
            }
        }
    }
}

int main() {
    scanf("%d%d%d%d%d", &m, &n, &s, &t1, &t2);
    for(int i = 1; i <= m; i++) {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
    }
    spfa(t1);
    minn = dis[s] + dis[t2];
    spfa(t2);
    minn = min(minn, dis[s]+dis[t1]);
    printf("%d", minn);
    return 0;
}

long long改回int，而後就A了。。

心裏一萬頭 * * * 奔過。。。。

不過幸虧最後仍是A了