题意：
给你n个人，每个人有一个权值 a_i ，（a_i是可以被100整除的)）现在需要你将n个人分成两组，有m个关系，a和b有关系代表a和b不能放在同一个组内，为了两组实力尽量平均，要你求两组权值差值最小时最大的值是哪一个
题解：二分图染色+dp
首先我们知道n个人必须全选分为两组，其次题目保证有解
因此我们很容易想到如果a->b,b->c，那么a一定和c要分在同一组内
这样我们就得到了很多个联通块
错误想法：我们得到了num个联通块后直接将num个联通块做01背包就可以求出差值最小时最大值，dp状态定义为前i个物品，容量为j时的最大值，但是实际上这样有可能把错误的情况考虑进来，例如两个矛盾的块同时放进一个背包，例如这组数据
2
4 1
300 300 100 500
1 2
6 4
1000 2000 1000 1500 1000 1500
1 2
2 3
4 5
5 6
01背包得到的答案是600 和 4000，实际上应该是 800和5000
正确想法，我们将物品分成联通块后，对分成的两个联通块做差值，之后枚举差值能否达到才是最优解，为了防止差值为负，我们在中间加上一个比较大的数即可
代码：错误写法：
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef long long ll;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
struct EDGE {
int v, nxt;
} edge[maxn << 1];
int head[maxn], tot;
void add_edge(int u, int v) {
edge[tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot++;
}
int a[maxn];
int num[maxn];
int num1, num2;
int vis[maxn];
int dp[205][maxn];
void dfs(int u, int flag) {
vis[u] = 1;
if(flag) num1 += a[u];
else num2 += a[u];
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(vis[v]) continue;
dfs(v, !flag);
}
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int T;
scanf("%d", &T);
while(T--) {
int n, m;
memset(head, -1, sizeof(head));
tot = 0;
int cur = 0;
int sum = 0, ret;
memset(vis, 0, sizeof(vis));
memset(num, 0, sizeof(num));
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
a[i] /= 100;
sum += a[i];
}
ret = sum / 2;
for(int i = 1, u, v; i <= m; i++) {
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
for(int i = 1; i <= n; i++) {
if(!vis[i]) {
num1 = 0, num2 = 0;
dfs(i, 0);
if(num1 != 0) {
num[++cur] = num1;
}
if(num2 != 0) {
num[++cur] = num2;
}
}
}
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= cur; i++) {
for(int j = 0; j <= ret; j++) {
dp[i][j] = dp[i - 1][j];
if(j >= num[i])dp[i][j] = max(dp[i][j], dp[i - 1][j - num[i]] + num[i]);
}
}
printf("%d\n", (sum - dp[cur][ret]) * 100);
}
return 0;
}
正确写法：
#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef long long ll;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const double eps = 1e-8;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const LL INFLL = 0x3f3f3f3f3f3f3f3f;
struct EDGE {
int v, nxt;
} edge[maxn << 1];
int head[maxn], tot;
void add_edge(int u, int v) {
edge[tot].v = v;
edge[tot].nxt = head[u];
head[u] = tot++;
}
int a[maxn];
int num[maxn];
int num1, num2;
int vis[maxn];
int dp[205][maxn];
void dfs(int u, int flag) {
vis[u] = 1;
if(flag) num1 += a[u];
else num2 += a[u];
for(int i = head[u]; i != -1; i = edge[i].nxt) {
int v = edge[i].v;
if(vis[v]) continue;
dfs(v, !flag);
}
}
int main() {
#ifndef ONLINE_JUDGE
FIN
#endif
int T;
scanf("%d", &T);
while(T--) {
int n, m;
memset(head, -1, sizeof(head));
tot = 0;
int cur = 0;
int sum = 0, ret;
memset(vis, 0, sizeof(vis));
memset(num, 0, sizeof(num));
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
a[i] /= 100;
sum += a[i];
}
for(int i = 1, u, v; i <= m; i++) {
scanf("%d%d", &u, &v);
add_edge(u, v);
add_edge(v, u);
}
ret = 0;
for(int i = 1; i <= n; i++) {
if(!vis[i]) {
num1 = 0, num2 = 0;
dfs(i, 0);
num[++cur] += abs (num1 - num2);
ret += abs(num1 - num2);
}
}
memset(dp, 0, sizeof(dp));
dp[0][100000] = 1;
for (int i = 1; i <= cur; i++) {
for (int j = -ret; j <= ret; j++) {
if (dp[i - 1][j - num[i] + 100000]) dp[i][j + 100000] = 1;
if (dp[i - 1][j + num[i] + 100000]) dp[i][j + 100000] = 1;
}
}
for (int i = 100000; i <= 100000 + ret; i++) {
if (dp[cur][i]) {
int x = i - 100000;
printf("%d\n", 100 * ((sum - x) / 2 + x));
break;
}
}
}
return 0;
}
每一个不曾刷题的日子 都是对生命的辜负 从弱小到强大，需要一段时间的沉淀，就是现在了 ~buerdepepeqi
