HDU 5294 Tricks Device (最大流+最短路)

题目链接：HDU 5294 Tricks Device

题意：n个点，m条边，并且一个人从1走到n只会走1到n的最短路径，问至少破坏几条边使原图的最短路不存在，最多破坏几条边使原图的最短路劲仍存在

思路：

1.跑一遍最短路，记录下所有最短路径，将这些最短路径的边以（0,1）（流量，容量）加到网络流中，跑一遍最大流

2.记录下的所有最短路径，再加到新的最短路的图中，边权为1，跑一遍最短路，m-最短路 就是答案

注意：有重边的情况

比如：

2 3

1 2 1

1 2 1

1 2 1

ans: 3 2

AC代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <set>
using namespace std;
const int MAXN=2010;
const int MAXM=60010;
#define typec int
const typec INF=0x3f3f3f3f;//防止后面溢出，这个不能太大

struct Edge {
	int to,next,cap,flow;
} edge[MAXM*4]; //注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void add(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];
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 init() {
	tol = 0;
	memset(head,-1,sizeof(head));
}

//-------------------------------------------------
struct Edge2 {
	int v;
	int cost;
	Edge2(int _v=0,int _cost=0):v(_v),cost(_cost){}
};
vector<Edge2>E[MAXN],E2[MAXN];
void addedge(int u,int v,int w) {
	E[u].push_back(Edge2(v,w));
}
bool vis[MAXN];//在队列标志
int cnt[MAXN];//每个点的入队列次数
int dist[MAXN];

bool SPFA(int start,int n) {
	memset(vis,false,sizeof(vis));
	for(int i=1;i<=n;i++)
		dist[i]=INF;
	vis[start]=true;
	dist[start]=0;
	queue<int>que;
	while(!que.empty())que.pop();
	que.push(start);
	memset(cnt,0,sizeof(cnt));
	cnt[start]=1;
	while(!que.empty())	{
		int u=que.front();
		que.pop();
		vis[u]=false;
		for(int i=0;i<E[u].size();i++){
			int v=E[u][i].v;
			if(dist[v]>dist[u]+E[u][i].cost){
				dist[v]=dist[u]+E[u][i].cost;
				if(!vis[v]){
					vis[v]=true;
					que.push(v);
					if(++cnt[v]>n) return false;
					//cnt[i]为入队列次数，用来判定是否存在负环回路
				}
			}
		}
	}
	return true;
}

set<int> s;
int main() {
	int i,j,m,n;
	int a,b,c;
	while(scanf("%d %d",&n,&m)!=EOF) {
		init();
		s.clear();
		for(i=1;i<=n;i++) E[i].clear();
		for(i=0; i<m; i++) {
			scanf("%d %d %d",&a,&b,&c);
			addedge(a,b,c);
			addedge(b,a,c);
		}
		int ok=SPFA(1,n);
		for(i=1;i<=n;i++){
			E2[i]=E[i];
			E[i].clear();
		}
		for(i=1; i<=n; i++) {
			int sz=E2[i].size();
			for(j=0; j<sz; j++) {
				int u=i;
				int v=E2[i][j].v;
				//得到最短路的边
				if(dist[u]+E2[i][j].cost==dist[v]) {
					add(u,v,1);
					s.insert(u),s.insert(v);
					addedge(u,v,1);
				}
			}
		}
		int ans=sap(1,n,(int)s.size());
		ok=SPFA(1,n);
		printf("%d %d\n",ans,m-dist[n]);
	}
	return 0;
}
/*
2 3
1 2 1
1 2 1
1 2 1
2 5
1 2 2
1 2 2
1 2 1
1 2 1
1 2 1
*/

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