标题效果：一个N积分m无向图边。它可以是路径k右边缘值变0,确定此时1-n最短路径长度。

Sol:我以为我们考虑分层图，图复制k+1部分，每间0~k一层。代表在这个时候已经过去“自由边缘”文章编号。

层与层之间的边权值为0且为单向由上层指向下层。

这样我们以0层的1点做单源最短路径。每一层的n点的距离最小值即为答案。

仅仅只是这种点数为O(K*N),边数为O（K*M）,比較慢。

我的做法是，对每一层使用heap-dijkstra算法由本层的原点更新这一层的最短路长度。然后显然能够用O(m)的复杂度推知下一层的初始最短路长度。

这样的做法显然空间和时间上均存在较大优势。

Code:

#include <cstdio>
#include <cstring>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
using namespace std;inline int getc() {static const int L = 1 << 15;static char buf[L], *S = buf, *T = buf;if (S == T) {T = (S = buf) + fread(buf, 1, L, stdin);if (S == T)return EOF;}return *S++;
}
inline int getint() {int c;while(!isdigit(c = getc()));int tmp = c - '0';while(isdigit(c = getc()))tmp = (tmp << 1) + (tmp << 3) + c - '0';return tmp;
}typedef long long LL;#define N 10010
#define M 50010
int n, m, k;
int head[N], next[M << 1], end[M << 1], len[M << 1];
LL dis[2][N];
bool inpath[N];queue<int> q;void addedge(int a, int b, int _len) {static int q = 1;len[q] = _len;end[q] = b;next[q] = head[a];head[a] = q++;
}
void make(int a, int b, int _len) {addedge(a, b, _len);addedge(b, a, _len);
}struct Node {int lab, dis;Node(int _lab = 0, int _dis = 0):lab(_lab),dis(_dis){}bool operator < (const Node &B) const {return (dis < B.dis) || (dis == B.dis && lab < B.lab);}
};
struct Heap {Node a[N];int top, ch[N];Heap():top(0){}void up(int x) {for(; x != 1; x >>= 1) {if (a[x] < a[x >> 1]) {swap(ch[a[x].lab], ch[a[x >> 1].lab]);swap(a[x], a[x >> 1]);}elsebreak;}}void down(int x) {int son;for(; x << 1 <= top; ) {son=(((x<<1)==top)||(a[x<<1]<a[(x<<1)|1]))?(x<<1):((x<<1)|1);if (a[son] < a[x]) {swap(ch[a[son].lab], ch[a[x].lab]);swap(a[son], a[x]);x = son;}elsebreak;}}void insert(Node x) {a[++top] = x;ch[x.lab] = top;up(top);}Node Min() {return a[1];}void pop() {a[1] = a[top];ch[a[top--].lab] = 1;down(1);}void change(int x, int to) {int ins = ch[x];a[ins].dis = to;up(ins);}
}H;void Dijkstra(bool d) {H.top = 0;int i, j;memset(inpath, 0, sizeof(inpath));for(i = 1; i <= n; ++i)H.insert(Node(i, dis[d][i]));for(i = 1; i <= n; ++i) {Node tmp = H.Min();H.pop();inpath[tmp.lab] = 1;for(j = head[tmp.lab]; j; j = next[j]) {if (!inpath[end[j]] && dis[d][end[j]] > dis[d][tmp.lab] + len[j]) {dis[d][end[j]] = dis[d][tmp.lab] + len[j];H.change(end[j], dis[d][end[j]]);}}}
}int main() {n = getint();m = getint();k = getint();int i, j;int a, b, x;for(i = 1; i <= m; ++i) {a = getint();b = getint();x = getint();make(a, b, x);}int now = 0, last = 1;memset(dis, 0x3f, sizeof(dis));dis[now][1] = 0;Dijkstra(now);LL ans = dis[now][n];while(k--) {now ^= 1;last ^= 1;for(i = 1; i <= n; ++i)dis[now][i] = dis[last][i];for(i = 1; i <= n; ++i)for(j = head[i]; j; j = next[j])dis[now][end[j]] = min(dis[now][end[j]], dis[last][i]);Dijkstra(now);if (ans == dis[now][n])break;ans = dis[now][n];}printf("%lld", ans);return 0;
}