POJ3255（spfa变形，很好的一道题）

题意：类似于次最短路，但却不是次最短路。此题要求的是比最短路大一点的路，意味着在从1到n的过程中夺走了一条边，我们现在求出多走一条边的最短路。

思路：既然是多走了一条边，那么遍历每一条边记录最小的值。题目保证一定存在这条边。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

using namespace std;

const int INF = 0x3f3f3f3f;
const int Nmax = 5005;
const int Mmax = 100005;

class Node
{
public:
int e,v;
int next;
}edge[Mmax*2];

int n,m;
int dis1[Nmax],dis2[Nmax];
int visit[Nmax];
int head[Nmax],pos;

void spfa(int s,int dist[])
{
for(int i = 1;i <= n; i++){
dist[i] = INF;
visit[i] = false;
}
queue<int>Q;
Q.push(s);
visit[s] = true;
dist[s] = 0;
while(!Q.empty()){
int u = Q.front();
Q.pop();
visit[u] = false;
for(int i = head[u];i != -1;i = edge[i].next){
int to = edge[i].e;
if(dist[u] + edge[i].v > dist[to])
continue;
dist[to] = dist[u] + edge[i].v;
if(!visit[to]){
Q.push(to);
visit[to] = true;
}
}
}
}

int main()
{
//freopen("in.txt","r",stdin);
scanf("%d%d",&n,&m);
memset(head,-1,sizeof(head));
pos = 0;
for(int i = 0;i < m; i++){
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
edge[pos].e = b;
edge[pos].v = c;
edge[pos].next = head[a];
head[a] = pos++;
edge[pos].e = a;
edge[pos].v = c;
edge[pos].next = head[b];
head[b] = pos++;
}
spfa(1,dis1);
spfa(n,dis2);
int ans = INF;
for(int i = 1;i <= n; i++){
for(int j = head[i];j != -1;j = edge[j].next){
int b = edge[j].e;
int c = edge[j].v;
int temp = dis1[i] + dis2[b] + c;
if(temp < ans && temp != dis1
)
ans = temp;
}
}
printf("%d\n",ans);
return 0;
}