codevs1403 新三国争霸 (最小生成树+dp)

这道题目的题意其实我觉得挺迷的，不过写过bzoj1003物流运输的话，这两道题考察的知识点类似，就可以明白什么意思了。

只不过bzoj1003是最短路+dp，而这道题是最小生成树+dp。

代码如下：

#include<iostream>
#include<cstdio>
#include<vector>
#include<queue>
#include<utility>
#include<stack>
#include<algorithm>
#include<cstring>
#include<string>
#include<cmath>
#include<set>
#include<map>
using namespace std;
const int maxn = 505;
const int INF = 0x3f3f3f3f;
int n, m, t, V, k;
int tot, par[maxn], cnt[maxn], sum[maxn], cost[55][55], dp[55];
bool vst[5005][55], flag[5005];

struct EDGE {
int u, v, val, id;
bool operator < (const EDGE& t) const {
if(val < t.val)
return 1;
return 0;
}
}edge[5005];

int Find(int x) {
int tmp = x;
while(x != par[x])
x = par[x];
int root = x;
x = tmp;
while(x != par[x]) {
tmp = par[x];
par[x] = root;
x = tmp;
}
return root;
}

void Union(int x, int y, int val) {
par[y] = x;
sum[x] += sum[y] + val;
cnt[x] += cnt[y];
}

void kruskal() {
for(int i = 1; i <= n; i++) {
par[i] = i;
sum[i] = 0;
cnt[i] = 1;
}
for(int i = 0; i < tot; i++) {
if(flag[edge[i].id])
continue;
int x = Find(edge[i].u);
int y = Find(edge[i].v);
if(x != y) {
Union(x, y, edge[i].val);
}
}
}

void build(int front, int back) {
memset(flag, 0, sizeof(flag));
for(int i = 0; i < tot; i++) {
for(int j = front; j <= back; j++) {
if(vst[i][j]) {
flag[i] = 1;
break;
}
}
}
}

int main() {
scanf("%d%d%d%d%d", &n, &m, &t, &V, &k);
while(m--) {
int u, v, val;
scanf("%d%d%d", &u, &v, &val);
if(u > v)
swap(u, v);
edge[tot].u = u;
edge[tot].v = v;
edge[tot].val = val;
edge[tot].id = tot;
tot++;
}
int p;
scanf("%d", &p);
while(p--) {
int u, v, t1, t2;
scanf("%d%d%d%d", &u, &v, &t1, &t2);
if(u > v)
swap(u, v);
for(int i = 0; i < tot; i++) {
if(edge[i].u == u && edge[i].v == v) {
for(int j = t1; j <= t2; j++) {
vst[i][j] = 1;
}
break;
}
}
}
sort(edge, edge + tot);
for(int i = 1; i <= t; i++) {
for(int j = i; j <= t; j++) {
build(i, j);
kruskal();
int root = Find(1);
if(cnt[root] == n)
cost[i][j] = sum[root];
else
cost[i][j] = INF;
}
}
fill(dp, dp + 55, INF);
dp[t + 1] = 0;
for(int i = t; i; i--) {
for(int j = i; j <= t; j++) {
dp[i] = min(dp[i], dp[j + 1] + cost[i][j] * (j - i + 1) * V + k);
}
}
printf("%d\n", dp[1]);
return 0;
}