SPOJ 962 Intergalactic Map (从A到B再到C的路线)
【题意】在一个无向图中,一个人要从A点赶往B点,之后再赶往C点,且要求中途不能多次经过同一个点。问是否存在这样的路线。(3 <= N <= 30011, 1 <= M <= 50011)
【思路】很巧的一道题,一般我们都是把源点连接起点,但那样的话就不好控制从A先到B再到C了,所以我们换个思路,以B为源点,A、C为汇点,看最大流是否为2即可~不经过同一个点就直接拆点连一条(i, i', 1)即可,无向图……就连两条反向边吧~~本来想改一下反向流就好的,可是想想那样也把源点汇点连出来的边也变成双向了……没试行不行……
#include
#include
#include
#include
#include
#include
#define MID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int MAXV = 60055;
const int MAXE = 200055;
const int oo = 0x3fffffff;
struct node{
int u, v, flow;
int opp;
int next;
};
struct Dinic{
node arc[2*MAXE];
int vn, en, head[MAXV]; //vn点个数(包括源点汇点),en边个数
int cur[MAXV]; //当前弧
int q[MAXV]; //bfs建层次图时的队列
int path[2*MAXE], top; //存dfs当前最短路径的栈
int dep[MAXV]; //各节点层次
void init(int n){
vn = n;
en = 0;
mem(head, -1);
}
void insert_flow(int u, int v, int flow){
arc[en].u = u;
arc[en].v = v;
arc[en].flow = flow;
arc[en].opp = en + 1;
arc[en].next = head[u];
head[u] = en ++;
arc[en].u = v;
arc[en].v = u;
arc[en].flow = 0; //反向弧
arc[en].opp = en - 1;
arc[en].next = head[v];
head[v] = en ++;
}
bool bfs(int s, int t){
mem(dep, -1);
int lq = 0, rq = 1;
dep[s] = 0;
q[lq] = s;
while(lq < rq){
int u = q[lq ++];
if (u == t){
return true;
}
for (int i = head[u]; i != -1; i = arc[i].next){
int v = arc[i].v;
if (dep[v] == -1 && arc[i].flow > 0){
dep[v] = dep[u] + 1;
q[rq ++] = v;
}
}
}
return false;
}
int solve(int s, int t){
int maxflow = 0;
while(bfs(s, t)){
int i, j;
for (i = 1; i <= vn; i ++) cur[i] = head[i];
for (i = s, top = 0;;){
if (i == t){
int mink;
int minflow = 0x3fffffff;
for (int k = 0; k < top; k ++)
if (minflow > arc[path[k]].flow){
minflow = arc[path[k]].flow;
mink = k;
}
for (int k = 0; k < top; k ++)
arc[path[k]].flow -= minflow, arc[arc[path[k]].opp].flow += minflow;
maxflow += minflow;
top = mink; //arc[mink]这条边流量变为0, 则直接回溯到该边的起点即可(这条边将不再包含在增广路内).
i = arc[path[top]].u;
}
for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
int v = arc[j].v;
if (arc[j].flow && dep[v] == dep[i] + 1)
break;
}
if (j != -1){
path[top ++] = j;
i = arc[j].v;
}
else{
if (top == 0) break;
dep[i] = -1;
i = arc[path[-- top]].u;
}
}
}
return maxflow;
}
}dinic;
int main(){
int t;
scanf("%d",&t);
while(t --){
int n, m;
scanf("%d %d", &n, &m);
if (n < 3){
puts("NO");
continue;
}
dinic.init(2*n+2);
for (int i = 1; i <= n; i ++){
dinic.insert_flow(2*i-1, 2*i, 1);
}
for (int i = 1; i <= m; i ++){
int u, v;
scanf("%d %d", &u, &v);
if(u > n || v > n) continue;
dinic.insert_flow(2*u, 2*v-1, 1);
dinic.insert_flow(2*v, 2*u-1, 1);
}
dinic.insert_flow(2*n+1, 2*2, 2);
dinic.insert_flow(2*1, 2*n+2, 1);
dinic.insert_flow(2*3, 2*n+2, 1);
if (dinic.solve(2*n+1, 2*n+2) == 2){
puts("YES");
}
else{
puts("NO");
}
}
return 0;
}
举杯独醉,饮罢飞雪,茫然又一年岁。 ------AbandonZHANG