标签：io for ar amp size ad har ef
/*刚开始不会写，最大点权独立集神马都不知道，在潘神的指导下终于做出来，灰常感谢ps；
和方格取数差不多奇偶建图，对于D必割点权为0，对于.必然不割点权为inf。然后和方格取数差不多的建图
.--.||E权值为2,，.||E--D权值为0.
最大点权独立集=sum-最小点权覆盖。
*/
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define inf 0x3fffffff
#define ii 50
#define N 3000
struct node {
int u,v,w,next;
}bian[N*6];
int head[N],yong,s,t,dis[N];
void init(){
yong=0;
memset(head,-1,sizeof(head));
memset(dis,-1,sizeof(dis));
}
void addedge(int u,int v,int w) {
bian[yong].u=u;
bian[yong].v=v;
bian[yong].w=w;
bian[yong].next=head[u];
head[u]=yong++;
}
void add(int u,int v,int w) {
addedge(u,v,w);
addedge(v,u,0);
}
void bfs() {
int u,v,i;
queue<int>q;
q.push(t);
dis[t]=0;
while(!q.empty()) {
u=q.front();
q.pop();
for(i=head[u];i!=-1;i=bian[i].next) {
v=bian[i].v;
if(dis[v]==-1) {
dis[v]=dis[u]+1;
q.push(v);
}
}
}
return ;
}
int ISAP() {
int sum=0;
bfs();
int gap[N],cur[N],stac[N],top,i;
memset(gap,0,sizeof(gap));
for(i=s;i<=t;i++) {
gap[dis[i]]++;
cur[i]=head[i];
}
int k=s;
top=0;
while(dis[s]<t+1) {
if(k==t) {
int minn=inf,index;
for(i=0;i<top;i++) {
int e=stac[i];
if(minn>bian[e].w) {
minn=bian[e].w;
index=i;
}
}
for(i=0;i<top;i++) {
int e=stac[i];
bian[e].w-=minn;
bian[e^1].w+=minn;
}
sum+=minn;
top=index;
k=bian[stac[top]].u;
}
for(i=cur[k];i!=-1;i=bian[i].next) {
int v=bian[i].v;
if(bian[i].w&&dis[k]==dis[v]+1) {
cur[k]=i;
k=v;
stac[top++]=i;
break;
}
}
if(i==-1) {
int m=t+1;
for(i=head[k];i!=-1;i=bian[i].next)
if(m>dis[bian[i].v]&&bian[i].w) {
m=dis[bian[i].v];
cur[k]=i;
}
if(--gap[dis[k]]==0)break;
gap[dis[k]=m+1]++;
if(k!=s)
k=bian[stac[--top]].u;
}
}
return sum;
}
int main() {
int n,m,i,j,T,id[ii][ii],cnt,sum,f=0;
char ss[ii][ii];
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
cnt=1;sum=0;
init();
for(i=1;i<=n;i++) {
scanf("%s",ss[i]+1);
for(j=1;j<=m;j++) {
if(ss[i][j]=='.'||ss[i][j]=='E')//记录总数
sum+=4;
id[i][j]=cnt++;
}
}
s=0;t=n*m+1;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++) {
if((i+j)&1) {//奇偶建图与源点相连
if(ss[i][j]=='.')
add(s,id[i][j],inf);//必然不割
if(ss[i][j]=='D')//必割
add(s,id[i][j],0);
if(ss[i][j]=='E')//任意选择
add(s,id[i][j],4);
if(i>=2) {
if(ss[i][j]=='.'||ss[i][j]=='E') {
if(ss[i-1][j]=='.'||ss[i-1][j]=='E')add(id[i][j],id[i-1][j],2);//
if(ss[i-1][j]=='D')add(id[i][j],id[i-1][j],0);//
}
else
add(id[i][j],id[i-1][j],0);
}
if(j>=2) {
if(ss[i][j]=='.'||ss[i][j]=='E') {
if(ss[i][j-1]=='.'||ss[i][j-1]=='E')add(id[i][j],id[i][j-1],2);
if(ss[i][j-1]=='D')add(id[i][j],id[i][j-1],0);
}
else
add(id[i][j],id[i][j-1],0);
}
if(i<=n-1){
if(ss[i][j]=='.'||ss[i][j]=='E') {
if(ss[i+1][j]=='.'||ss[i+1][j]=='E')add(id[i][j],id[i+1][j],2);
if(ss[i+1][j]=='D')add(id[i][j],id[i+1][j],0);
}
else
add(id[i][j],id[i+1][j],0);
}
if(j<=m-1) {
if(ss[i][j]=='.'||ss[i][j]=='E') {
if(ss[i][j+1]=='.'||ss[i][j+1]=='E')add(id[i][j],id[i][j+1],2);
if(ss[i][j+1]=='D')add(id[i][j],id[i][j+1],0);
}
else
add(id[i][j],id[i][j+1],0);
}
}
else {//与汇点相连
if(ss[i][j]=='.')
add(id[i][j],t,inf);
if(ss[i][j]=='E')
add(id[i][j],t,4);
if(ss[i][j]=='D')
add(id[i][j],t,0);
}
}
printf("Case %d: ",++f);
printf("%d\n",sum-ISAP());
}
return 0;
}
hdu 4859 最大点权独立集的变形（方格取数的变形）,布布扣,bubuko.com
标签：io for ar amp size ad har ef
原文地址：http://blog.csdn.net/u011483306/article/details/38406677
