题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5294
Problem Description
Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance of the tomb while Dumb Zhang’s at the end of it. The tomb is made up of many chambers, the total number is N. And there are M channels connecting the chambers. Innocent Wu wants
to catch up Dumb Zhang to find out the answers of some questions, however, it’s Dumb Zhang’s intention to keep Innocent Wu in the dark, to do which he has to stop Innocent Wu from getting him. Only via the original shortest ways from the entrance to the end
of the tomb costs the minimum time, and that’s the only chance Innocent Wu can catch Dumb Zhang.
Unfortunately, Dumb Zhang masters the art of becoming invisible(奇门遁甲) and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.
Unfortunately, Dumb Zhang masters the art of becoming invisible(奇门遁甲) and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.
Input
There are multiple test cases. Please process till EOF.
For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.
In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0<li<=100) minute to pass channel i.
The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.
For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.
In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0<li<=100) minute to pass channel i.
The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.
Output
Output two numbers to stand for the answers of Dumb Zhang and Innocent Wu’s questions.
Sample Input
8 9 1 2 2 2 3 2 2 4 1 3 5 3 4 5 4 5 8 1 1 6 2 6 7 5 7 8 1
Sample Output
2 6
Author
FZUACM
Source
转
题意：
给出n个墓室，m条路径。一个人在1号墓室（起点）。还有一个人在n号墓室（终点）；
起点的那个人仅仅有通过最短路径才干追上终点的那个人。而终点的那个人能切断随意路径。
第一问——终点那人要使起点那人不能追上的情况下能够切的最少的路径数。输出最少的路径数
第二问——起点那人能追上终点那人的情况下，终点那人能切断的最多的路径数。输出最多的路径数
PS:先跑最短路。
然后通过 dist[i]-dist[j] == map[j][i]
假设符合的话 map[j][i]就是 最短路中的一条边。
然后把这些最短路的边 建图。跑最大流。流量是有多少边权同样的重边，跑出来就是最小割，也就是阻断全部最短路的最小花费。花费是每破坏一条路为1。所以出来的值。就是破坏了多少的边。
然后如最大流相同的建边，跑最短路，边权为1，跑出来的最短路dist[n]，就是 跨越边数最少的 最短路的边数了。
官方题解：
代码例如以下：
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define MAXN 800047//点数的最大值
#define MAXM 2247//边数的最大值
int head[MAXM], pre[MAXM];
int dep[MAXM], cur[MAXM], gap[MAXM];//gap[x]=y :说明残留网络中dep[i]==x的个数为y
int EN;
struct Edge
{
int to,next,cap,flow;
} edge[MAXN]; //注意是MAXM
int tol;
int k, c, m;
int s, e;//源点。汇点
int map[MAXM][MAXM];
int cost1[MAXM][MAXM], cost2[MAXM][MAXM];
int num[MAXM][MAXM];//记录边数。>1 既有重边
//加边，单向图三个參数，双向图四个參数
void addedge(int u,int v,int w,int rw = 0)
{
edge[tol].to = v;
edge[tol].cap = w;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = rw;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
int Q[MAXN];
void BFS(int start,int end)
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0] = 1;
int front = 0, rear = 0;
dep[end] = 0;
Q[rear++] = end;
while(front != rear)
{
int u = Q[front++];
for(int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if(dep[v] != -1)continue;
Q[rear++] = v;
dep[v] = dep[u] + 1;
gap[dep[v]]++;
}
}
}
int S[MAXN];
//输入參数：起点、终点、点的总数
//点的编号没有影响。仅仅要输入点的总数
int sap(int start,int end,int N)
{
BFS(start,end);
memcpy(cur,head,sizeof(head));
int top = 0;
int u = start;
int ans = 0;
while(dep[start] < N)
{
if(u == end)
{
int Min = INF;
int inser;
for(int i = 0; i < top; i++)
if(Min > edge[S[i]].cap - edge[S[i]].flow)
{
Min = edge[S[i]].cap - edge[S[i]].flow;
inser = i;
}
for(int i = 0; i < top; i++)
{
edge[S[i]].flow += Min;
edge[S[i]^1].flow -= Min;
}
ans += Min;
top = inser;
u = edge[S[top]^1].to;
continue;
}
bool flag = false;
int v;
for(int i = cur[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
{
flag = true;
cur[u] = i;
break;
}
}
if(flag)
{
S[top++] = cur[u];
u = v;
continue;
}
int Min = N;
for(int i = head[u]; i != -1; i = edge[i].next)
if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if(u != start)u = edge[S[--top]^1].to;
}
return ans;
}
void Dijkstra(int s, int n, int cost[][MAXM], int dis[MAXM])
{
//int dis[MAXM];//记录到随意点的最短距离
int mark[MAXM];//记录被选中的结点
int i, j, k;
for(i = 1; i <= n; i++)
{
mark[i] = 0;//初始化全部结点。每一个结点都没有被选中
dis[i] = INF;
}
mark[s] = 1;//start结点被选中
dis[s] = 0;//将start结点的的距离设置为0
int MIN;//设置最短的距离。
for(i = 1; i <= n; i++)
{
k = 1;//赋初值非常重要
MIN = INF;
for(j = 1; j <= n; j++)
{
if(!mark[j] && dis[j] < MIN)//未被选中的结点中，距离最短的被选中
{
MIN = dis[j] ;
k = j;
}
}
mark[k] = 1;//标记为被选中
for(j = 1; j <= n; j++)
{
if(!mark[j] && dis[j]>dis[k] + cost[k][j])//改动剩余结点的最短距离
{
dis[j] = dis[k] + cost[k][j];
}
}
}
}
void init()
{
memset(head,-1,sizeof(head));
memset(cost1,INF,sizeof(cost1));
memset(cost2,INF,sizeof(cost2));
memset(num,0,sizeof(num));
EN = 0;
}
int main()
{
int n, m;
int u, v, w;
int dis1[MAXM], dis2[MAXM];
while(~scanf("%d%d",&n,&m))
{
init();
for(int i = 1; i <= n; i++)
{
cost1[i][i] = 0;
}
for(int i = 1; i <= m; i++)
{
scanf("%d%d%d",&u,&v,&w);
if(cost1[u][v] > w)
{
cost1[u][v] = w;
cost1[v][u] = w;
num[u][v] = 1;
num[v][u] = 1;
}
else if(cost1[u][v] == w)//重边
{
num[u][v]++;
num[v][u]++;
}
}
Dijkstra(1,n,cost1,dis1);
for(int i = 1; i <= n; i++)
{
cost2[i][i] = 0;
}
for(int i = 1; i <= n; i++)
{
for(int j = 1; j <= n; j++)
{
if(i != j)
{
if(dis1[i] - dis1[j] == cost1[i][j])//最短路的边
{
cost2[j][i] = 1;
addedge(j,i,num[i][j]);
}
}
}
}
Dijkstra(1,n,cost2,dis2);
printf("%d %d
",sap(1,n,n),m-dis2[n]);
}
return 0;
}