Luogu 2865 [USACO06NOV]路障Roadblocks

题目描述

贝茜把家搬到了一个小农场，但她常常回到FJ的农场去拜访她的朋友。贝茜很喜欢路边的风景，不想那么快地结束她的旅途，于是她每次回农场，都会选择第二短的路径，而不象我们所习惯的那样，选择最短路。 贝茜所在的乡村有R(1<=R<=100,000)条双向道路，每条路都联结了所有的N(1<=N<=5000)个农场中的某两个。贝茜居住在农场1，她的朋友们居住在农场N（即贝茜每次旅行的目的地）。 贝茜选择的第二短的路径中，可以包含任何一条在最短路中出现的道路，并且，一条路可以重复走多次。当然咯，第二短路的长度必须严格大于最短路（可能有多条）的长度，但它的长度必须不大于所有除最短路外的路径的长度。

输入输出格式

输入格式：
Line 1: Two space-separated integers: N and R
Lines 2..R+1: Each line contains three space-separated integers: A, B, and D that describe a road that connects intersections A and B and has length D (1 ≤ D ≤ 5000)

输出格式：
Line 1: The length of the second shortest path between node 1 and node N

代码

这道题是次短路的题，我们可以用spfa来求解
需要开两个数组dist1[]记录最短路，dist2[]记录次短路
dist1[v]>dist1[u]+w(u, v) 显然要更新dist1，而因为dist1是最短路，所以显然要先更新dist2等于原来的dist1再更新dist1
dist2[v]>dis1t[u]+w(u, v) && dist1[v]>dist1[u]+w(u, v) 因为现在 d1[u]+w(u, v) 不是最短路，但是又比次短路小，那么显然要更新次短路
dist2[v]>dist2[u]+w(u, v) 这时次短路比次短路短，那么肯定要更新次短路.

A C历程

一开始只想到上面的第一种和第二种情况，没有考虑第三种情况，所以只有50分，所有没AC的点的数据都要大些。

然后我参考了hzwer大神的题解知道了第三种情况，但我将次短路也初始化为零了，所以每个点的次短路都变成了最短了，于是我就在判断第三种情况下加了判断dist1[v]

#include<cstdio>#include<iostream>#include<algorithm>#include<deque>#include<cstring>const int maxn = 5005;using namespace std;struct edge{    int v;    int next;    int w;}e[200050];int ans,dist[maxn],p[maxn],dist2[maxn],head[maxn],cnt;bool vis[maxn];void add(int u,int v,int w){    e[++cnt] = (edge){v,head[u],w};    head[u] = cnt;}int read() //读入优化 {    int ans=0;char ch = getchar();    while(ch<'0'||ch>'9')    ch = getchar();    while(ch>='0'&&ch<='9') {ans = ans *10 + ch - '0';ch = getchar();}    return ans;}void spfa(){    int v,u;    deque<int> q;    memset(dist,0x7f,sizeof(dist));    memset(vis,false,sizeof(vis));    q.push_back(1);    dist[1] = 0;    vis[1] = true;    while(!q.empty())    {        u = q.front();        q.pop_front();        for(int i=head[u];i;i=e[i].next)        {            v = e[i].v;            if(dist[v]>dist[u]+e[i].w)            {                dist2[v] = dist[v];                 dist[v] = dist[u] + e[i].w;                if(!vis[v])                {                    q.push_back(v);                    vis[v] = true;                }            }            else if(dist2[v]>dist[u]+e[i].w)            {                dist2[v] = dist[u] + e[i].w;            }        }        vis[u] = false;    }}int main(){    int n,m;    n=read();    m=read();     for(int i=1;i<=m;i++)    {        int u,v,w;        u=read();        v=read();        w=read();        add(u,v,w);        add(v,u,w);    }    spfa();    ans = dist2[n];    printf("%d",ans);    return 0;}

90分

#include<cstdio>#include<iostream>#include<algorithm>#include<deque>#include<cstring>const int maxn = 5005;using namespace std;struct edge{    int v;    int next;    int w;}e[200050];int ans,dist[maxn],p[maxn],dist2[maxn],head[maxn],cnt;bool vis[maxn];void add(int u,int v,int w){    e[++cnt] = (edge){v,head[u],w};    head[u] = cnt;}int read() //读入优化 {    int ans=0;char ch = getchar();    while(ch<'0'||ch>'9')    ch = getchar();    while(ch>='0'&&ch<='9') {ans = ans *10 + ch - '0';ch = getchar();}    return ans;}void spfa(){    int v,u;    deque<int> q;    memset(dist,0x7f,sizeof(dist));    memset(dist2,0x7f,sizeof(dist2));    memset(vis,false,sizeof(vis));    q.push_back(1);    dist[1] = 0;    dist2[1] = 0;    vis[1] = true;    while(!q.empty())    {        u = q.front();        q.pop_front();        for(int i=head[u];i;i=e[i].next)        {            v = e[i].v;            if(dist[v]>dist[u]+e[i].w)            {                dist2[v] = dist[v];                 dist[v] = dist[u] + e[i].w;                if(!vis[v])                {                    q.push_back(v);                    vis[v] = true;                }            }            else if(dist2[v]>dist[u]+e[i].w&&dist[v]!=dist[u]+e[i].w)            {                dist2[v] = dist[u] + e[i].w;                if(!vis[v])                {                    q.push_back(v);                    vis[v] = true;                }            }            if(dist2[u]+e[i].w<dist2[v]&&dist[v]<dist2[u]+e[i].w)            {                dist2[v] = dist2[u] + e[i].w;                if(!vis[v])                {                    q.push_back(v);                    vis[v] = true;                }            }        }        vis[u] = false;    }}int main(){    int n,m;    n=read();    m=read();     for(int i=1;i<=m;i++)    {        int u,v,w;        u=read();        v=read();        w=read();        add(u,v,w);        add(v,u,w);    }    spfa();    ans = dist2[n];    printf("%d",ans);    return 0;}

100分

#include<cstdio>#include<iostream>#include<algorithm>#include<deque>#include<cstring>const int maxn = 5005;using namespace std;struct edge{    int v;    int next;    int w;}e[200050];int ans,dist[maxn],p[maxn],dist2[maxn],head[maxn],cnt;bool vis[maxn];void add(int u,int v,int w){    e[++cnt] = (edge){v,head[u],w};    head[u] = cnt;}int read() //读入优化 {    int ans=0;char ch = getchar();    while(ch<'0'||ch>'9')    ch = getchar();    while(ch>='0'&&ch<='9') {ans = ans *10 + ch - '0';ch = getchar();}    return ans;}void spfa(){    int v,u;    deque<int> q;    memset(dist,0x7f,sizeof(dist));    memset(dist2,0x7f,sizeof(dist2));    memset(vis,false,sizeof(vis));    q.push_back(1);    dist[1] = 0;    vis[1] = true;    while(!q.empty())    {        u = q.front();        q.pop_front();        for(int i=head[u];i;i=e[i].next)        {            v = e[i].v;            if(dist[v]>dist[u]+e[i].w)            {                dist2[v] = dist[v];                 dist[v] = dist[u] + e[i].w;                if(!vis[v])                {                    q.push_back(v);                    vis[v] = true;                }            }            if(dist2[v] > dist[u]+e[i].w && dist[v] < dist[u]+e[i].w)            {                dist2[v] = dist[u] + e[i].w;                if(!vis[v])                {                    q.push_back(v);                    vis[v] = true;                }            }            if(dist2[u]+e[i].w<dist2[v])            {                dist2[v] = dist2[u] + e[i].w;                if(!vis[v])                {                    q.push_back(v);                    vis[v] = true;                }            }        }        vis[u] = false;    }}int main(){    int n,m;    n=read();    m=read();     for(int i=1;i<=m;i++)    {        int u,v,w;        u=read();        v=read();        w=read();        add(u,v,w);        add(v,u,w);    }    spfa();    ans = dist2[n];    printf("%d",ans);    return 0;}